[proofplan] We compute both kernels from orthonormal monomial bases. For the ball, the only nontrivial point is the exact norm of each monomial; we obtain it by comparing a product [Gaussian integral](/theorems/1140) on $\mathbb{C}^n$ with polar coordinates. Summing the resulting basis expansion gives the binomial series in the single scalar variable $\langle z,w\rangle$. For the polydisc, the same monomial method factors coordinate-by-coordinate and gives the product of the one-variable disc series. [/proofplan] [step:Fix the Bergman space convention and the monomial notation] For a bounded domain $\Omega \subset \mathbb{C}^n$, define \begin{align*} A^2(\Omega) := \left\{ f:\Omega \to \mathbb{C} : f \text{ is holomorphic and } \int_\Omega |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta)<\infty \right\}. \end{align*} We equip $A^2(\Omega)$ with the Hilbert-space inner product \begin{align*} (f,g)_{A^2(\Omega)} := \int_\Omega f(\zeta)\overline{g(\zeta)}\,d\mathcal{L}^{2n}(\zeta). \end{align*} For the domains used below, the point-evaluation maps are bounded. Indeed, if $\Omega$ is either $B_n$ or $\Delta^n$ and $z\in\Omega$, choose $\rho>0$ such that the Euclidean ball $B(z,\rho):=\{\zeta\in\mathbb{C}^n:|\zeta-z|<\rho\}$ is contained in $\Omega$. The sub-mean-value inequality for the nonnegative subharmonic function $|f|^2$ gives \begin{align*} |f(z)|^2 \leq \frac{1}{\mathcal{L}^{2n}(B(z,\rho))} \int_{B(z,\rho)}|f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \frac{1}{\mathcal{L}^{2n}(B(z,\rho))} \|f\|_{A^2(\Omega)}^2. \end{align*} Hence the [Riesz representation theorem](/theorems/221) applies to each evaluation functional $E_z:A^2(\Omega)\to\mathbb{C}$, $E_z(f):=f(z)$, and defines the Bergman kernel vector $K_\Omega(\cdot,z)\in A^2(\Omega)$. The Bergman kernel $K_\Omega:\Omega \times \Omega \to \mathbb{C}$ is characterized by being holomorphic in the first variable, antiholomorphic in the second variable, and satisfying \begin{align*} f(z) = (f,K_\Omega(\cdot,z))_{A^2(\Omega)} \end{align*} for every $f \in A^2(\Omega)$ and every $z \in \Omega$. Let $\mathbb{N}_0:=\mathbb{N}\cup\{0\}$ denote the set of nonnegative integers, and let \begin{align*} \Delta:=\{\zeta\in\mathbb{C}:|\zeta|<1\} \end{align*} denote the one-dimensional unit disc. We write $\mathcal{L}^1$ for Lebesgue measure on $\mathbb{R}$, $\mathcal{L}^2$ for Lebesgue measure on $\mathbb{C}\simeq\mathbb{R}^2$, and $\mathcal{L}^{2n}$ for Lebesgue measure on $\mathbb{C}^n\simeq\mathbb{R}^{2n}$. For a multi-index $\alpha=(\alpha_1,\dots,\alpha_n)\in \mathbb{N}_0^n$, define \begin{align*} |\alpha|&:=\alpha_1+\cdots+\alpha_n,\\ \alpha!&:=\alpha_1!\cdots \alpha_n!,\\ z^\alpha&:=z_1^{\alpha_1}\cdots z_n^{\alpha_n} \end{align*} for $z=(z_1,\dots,z_n)\in\mathbb{C}^n$. The monomials are mutually orthogonal on both $B_n$ and $\Delta^n$. Indeed, both domains are invariant under the coordinatewise rotations \begin{align*} R_\theta:\mathbb{C}^n&\to\mathbb{C}^n\\ (z_1,\dots,z_n)&\mapsto (e^{i\theta_1}z_1,\dots,e^{i\theta_n}z_n), \end{align*} where $\theta=(\theta_1,\dots,\theta_n)\in\mathbb{R}^n$, and $\mathcal{L}^{2n}$ is invariant under $R_\theta$. Hence, if $\alpha\neq\beta$, choose $j$ with $\alpha_j\neq\beta_j$. Then \begin{align*} \int_\Omega z^\alpha \overline{z^\beta}\,d\mathcal{L}^{2n}(z) &= e^{i(\alpha_j-\beta_j)\theta_j} \int_\Omega z^\alpha \overline{z^\beta}\,d\mathcal{L}^{2n}(z) \end{align*} for every $\theta_j\in\mathbb{R}$, forcing the integral to vanish. [guided] We first make the Hilbert-space convention explicit because the kernel formula depends on the convention for the inner product. For a bounded domain $\Omega \subset \mathbb{C}^n$, the Bergman space is \begin{align*} A^2(\Omega) := \left\{ f:\Omega \to \mathbb{C} : f \text{ is holomorphic and } \int_\Omega |f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta)<\infty \right\}. \end{align*} Its inner product is \begin{align*} (f,g)_{A^2(\Omega)} := \int_\Omega f(\zeta)\overline{g(\zeta)}\,d\mathcal{L}^{2n}(\zeta). \end{align*} With this convention, point evaluations are bounded on the two domains used in the theorem. If $\Omega$ is either $B_n$ or $\Delta^n$ and $z\in\Omega$, choose $\rho>0$ such that the Euclidean ball $B(z,\rho):=\{\zeta\in\mathbb{C}^n:|\zeta-z|<\rho\}$ is contained in $\Omega$. Since $|f|^2$ is subharmonic for holomorphic $f$, the sub-mean-value inequality gives \begin{align*} |f(z)|^2 \leq \frac{1}{\mathcal{L}^{2n}(B(z,\rho))} \int_{B(z,\rho)}|f(\zeta)|^2\,d\mathcal{L}^{2n}(\zeta) \leq \frac{1}{\mathcal{L}^{2n}(B(z,\rho))} \|f\|_{A^2(\Omega)}^2. \end{align*} Thus $E_z:A^2(\Omega)\to\mathbb{C}$, $E_z(f):=f(z)$, is a bounded linear functional. By the [Riesz representation theorem](/theorems/218), there is a unique vector $K_\Omega(\cdot,z)\in A^2(\Omega)$ representing it. The Bergman kernel is holomorphic in its first variable and antiholomorphic in its second variable, and it satisfies \begin{align*} f(z) = (f,K_\Omega(\cdot,z))_{A^2(\Omega)}. \end{align*} This order of variables matters: the inner product is linear in the first slot and conjugate-linear in the second slot. The proof will use monomials. Let $\mathbb{N}_0:=\mathbb{N}\cup\{0\}$ be the set of nonnegative integers, and let $\Delta:=\{\zeta\in\mathbb{C}:|\zeta|<1\}$ be the one-dimensional unit disc. We write $\mathcal{L}^1$ for Lebesgue measure on $\mathbb{R}$, $\mathcal{L}^2$ for Lebesgue measure on $\mathbb{C}\simeq\mathbb{R}^2$, and $\mathcal{L}^{2n}$ for Lebesgue measure on $\mathbb{C}^n\simeq\mathbb{R}^{2n}$. For $\alpha=(\alpha_1,\dots,\alpha_n)\in \mathbb{N}_0^n$, define \begin{align*} |\alpha|&:=\alpha_1+\cdots+\alpha_n,\\ \alpha!&:=\alpha_1!\cdots \alpha_n!,\\ z^\alpha&:=z_1^{\alpha_1}\cdots z_n^{\alpha_n}. \end{align*} The key structural fact is that different monomials are orthogonal. The reason is rotational symmetry. Let \begin{align*} R_\theta:\mathbb{C}^n&\to\mathbb{C}^n\\ (z_1,\dots,z_n)&\mapsto (e^{i\theta_1}z_1,\dots,e^{i\theta_n}z_n) \end{align*} be the coordinatewise rotation. Both $B_n$ and $\Delta^n$ are invariant under $R_\theta$, and Lebesgue measure $\mathcal{L}^{2n}$ is invariant under this unitary [linear map](/page/Linear%20Map). Therefore, for $\Omega=B_n$ or $\Omega=\Delta^n$, \begin{align*} \int_\Omega z^\alpha \overline{z^\beta}\,d\mathcal{L}^{2n}(z) &= \int_\Omega (R_\theta z)^\alpha \overline{(R_\theta z)^\beta}\,d\mathcal{L}^{2n}(z)\\ &= e^{i\sum_{j=1}^n(\alpha_j-\beta_j)\theta_j} \int_\Omega z^\alpha \overline{z^\beta}\,d\mathcal{L}^{2n}(z). \end{align*} If $\alpha\neq\beta$, choose $j$ with $\alpha_j\neq\beta_j$ and vary $\theta_j$. The only complex number equal to all of its nontrivial rotations is $0$, so the integral is $0$. [/guided] [/step] [step:Compute the exact $A^2(B_n)$ norm of each monomial] Let $\sigma$ denote the surface measure $\mathcal{H}^{2n-1}$ on the unit sphere \begin{align*} S^{2n-1}:=\{\xi\in\mathbb{C}^n:|\xi|=1\}. \end{align*} For $\alpha\in\mathbb{N}_0^n$, define \begin{align*} I_\alpha := \int_{\mathbb{C}^n}|z^\alpha|^2e^{-|z|^2}\,d\mathcal{L}^{2n}(z). \end{align*} Writing each coordinate $z_j\in\mathbb{C}$ in one-dimensional polar form gives \begin{align*} I_\alpha &= \prod_{j=1}^n \int_{\mathbb{C}} |z_j|^{2\alpha_j}e^{-|z_j|^2}\,d\mathcal{L}^2(z_j)\\ &= \prod_{j=1}^n \left( 2\pi\int_0^\infty r^{2\alpha_j+1}e^{-r^2}\,d\mathcal{L}^1(r) \right)\\ &= \prod_{j=1}^n \pi \alpha_j! = \pi^n\alpha!. \end{align*} On the other hand, using polar coordinates $z=r\xi$ in $\mathbb{C}^n\simeq\mathbb{R}^{2n}$ gives \begin{align*} I_\alpha &= \int_0^\infty \int_{S^{2n-1}} r^{2|\alpha|}|\xi^\alpha|^2e^{-r^2}r^{2n-1} \,d\sigma(\xi)\,d\mathcal{L}^1(r)\\ &= \left(\int_{S^{2n-1}}|\xi^\alpha|^2\,d\sigma(\xi)\right) \left(\int_0^\infty r^{2|\alpha|+2n-1}e^{-r^2}\,d\mathcal{L}^1(r)\right)\\ &= \left(\int_{S^{2n-1}}|\xi^\alpha|^2\,d\sigma(\xi)\right) \frac{(|\alpha|+n-1)!}{2}. \end{align*} Thus \begin{align*} \int_{S^{2n-1}}|\xi^\alpha|^2\,d\sigma(\xi) = \frac{2\pi^n\alpha!}{(|\alpha|+n-1)!}. \end{align*} Applying polar coordinates on the unit ball, \begin{align*} \int_{B_n}|z^\alpha|^2\,d\mathcal{L}^{2n}(z) &= \int_0^1 \int_{S^{2n-1}} r^{2|\alpha|}|\xi^\alpha|^2r^{2n-1} \,d\sigma(\xi)\,d\mathcal{L}^1(r)\\ &= \frac{1}{2(|\alpha|+n)} \int_{S^{2n-1}}|\xi^\alpha|^2\,d\sigma(\xi)\\ &= \frac{\pi^n\alpha!}{(|\alpha|+n)!}. \end{align*} Therefore \begin{align*} \left\|z^\alpha\right\|_{A^2(B_n)}^2 = \frac{\pi^n\alpha!}{(n+|\alpha|)!}. \end{align*} [guided] The main computation for the ball is the exact norm of $z^\alpha$. We compute it indirectly through a [Gaussian integral](/theorems/1140), because the [Gaussian integral](/theorems/1140) factors over the coordinates while polar coordinates reveal the spherical average needed for the ball. Let $\sigma$ be the surface measure $\mathcal{H}^{2n-1}$ on \begin{align*} S^{2n-1}:=\{\xi\in\mathbb{C}^n:|\xi|=1\}. \end{align*} For $\alpha\in\mathbb{N}_0^n$, define \begin{align*} I_\alpha := \int_{\mathbb{C}^n}|z^\alpha|^2e^{-|z|^2}\,d\mathcal{L}^{2n}(z). \end{align*} First compute $I_\alpha$ by separating variables. Since \begin{align*} |z^\alpha|^2e^{-|z|^2} = \prod_{j=1}^n |z_j|^{2\alpha_j}e^{-|z_j|^2}, \end{align*} Tonelli's theorem applies because the integrand is nonnegative, and hence \begin{align*} I_\alpha &= \prod_{j=1}^n \int_{\mathbb{C}} |z_j|^{2\alpha_j}e^{-|z_j|^2}\,d\mathcal{L}^2(z_j). \end{align*} In one complex variable, write $z_j=re^{i\theta}$ with $r\in(0,\infty)$ and $\theta\in[0,2\pi)$. The planar Lebesgue measure transforms as $d\mathcal{L}^2(z_j)=r\,d\mathcal{L}^1(r)\,d\mathcal{L}^1(\theta)$, so \begin{align*} \int_{\mathbb{C}} |z_j|^{2\alpha_j}e^{-|z_j|^2}\,d\mathcal{L}^2(z_j) &= 2\pi\int_0^\infty r^{2\alpha_j+1}e^{-r^2}\,d\mathcal{L}^1(r). \end{align*} With the substitution $s=r^2$, so that $ds=2r\,d\mathcal{L}^1(r)$, the last integral becomes \begin{align*} 2\pi\int_0^\infty r^{2\alpha_j+1}e^{-r^2}\,d\mathcal{L}^1(r) &= \pi\int_0^\infty s^{\alpha_j}e^{-s}\,d\mathcal{L}^1(s) = \pi\alpha_j!. \end{align*} Therefore \begin{align*} I_\alpha=\pi^n\alpha!. \end{align*} Now compute the same integral in polar coordinates on $\mathbb{C}^n\simeq\mathbb{R}^{2n}$. The substitution is $z=r\xi$, where $r\in(0,\infty)$ and $\xi\in S^{2n-1}$, and Lebesgue measure decomposes as \begin{align*} d\mathcal{L}^{2n}(z)=r^{2n-1}\,d\sigma(\xi)\,d\mathcal{L}^1(r). \end{align*} Since $(r\xi)^\alpha=r^{|\alpha|}\xi^\alpha$, we get \begin{align*} I_\alpha &= \int_0^\infty \int_{S^{2n-1}} r^{2|\alpha|}|\xi^\alpha|^2e^{-r^2}r^{2n-1} \,d\sigma(\xi)\,d\mathcal{L}^1(r)\\ &= \left(\int_{S^{2n-1}}|\xi^\alpha|^2\,d\sigma(\xi)\right) \left(\int_0^\infty r^{2|\alpha|+2n-1}e^{-r^2}\,d\mathcal{L}^1(r)\right). \end{align*} Again using $s=r^2$, the radial integral is \begin{align*} \int_0^\infty r^{2|\alpha|+2n-1}e^{-r^2}\,d\mathcal{L}^1(r) &= \frac{1}{2}\int_0^\infty s^{|\alpha|+n-1}e^{-s}\,d\mathcal{L}^1(s)\\ &= \frac{(|\alpha|+n-1)!}{2}. \end{align*} Comparing this with $I_\alpha=\pi^n\alpha!$ gives \begin{align*} \int_{S^{2n-1}}|\xi^\alpha|^2\,d\sigma(\xi) = \frac{2\pi^n\alpha!}{(|\alpha|+n-1)!}. \end{align*} Finally, integrate over the unit ball. The same polar-coordinate substitution now has $0<r<1$, because $z\in B_n$ exactly when $z=r\xi$ with $0<r<1$ and $\xi\in S^{2n-1}$. Hence \begin{align*} \int_{B_n}|z^\alpha|^2\,d\mathcal{L}^{2n}(z) &= \int_0^1 \int_{S^{2n-1}} r^{2|\alpha|}|\xi^\alpha|^2r^{2n-1} \,d\sigma(\xi)\,d\mathcal{L}^1(r)\\ &= \left(\int_{S^{2n-1}}|\xi^\alpha|^2\,d\sigma(\xi)\right) \left(\int_0^1 r^{2|\alpha|+2n-1}\,d\mathcal{L}^1(r)\right)\\ &= \frac{2\pi^n\alpha!}{(|\alpha|+n-1)!}\cdot \frac{1}{2(|\alpha|+n)}\\ &= \frac{\pi^n\alpha!}{(|\alpha|+n)!}. \end{align*} Thus \begin{align*} \left\|z^\alpha\right\|_{A^2(B_n)}^2 = \frac{\pi^n\alpha!}{(n+|\alpha|)!}. \end{align*} [/guided] [/step] [step:Sum the normalized monomial expansion for the ball] Define, for $\alpha\in\mathbb{N}_0^n$, \begin{align*} e_\alpha:B_n&\to\mathbb{C}\\ z&\mapsto \left(\frac{(n+|\alpha|)!}{\pi^n\alpha!}\right)^{1/2}z^\alpha. \end{align*} The preceding step and orthogonality show that $(e_\alpha)_{\alpha\in\mathbb{N}_0^n}$ is an orthonormal sequence in $A^2(B_n)$. To prove completeness, let $f\in A^2(B_n)$ be orthogonal to every monomial, and write its Taylor expansion at $0$ as \begin{align*} f(z)=\sum_{\alpha\in\mathbb{N}_0^n}a_\alpha z^\alpha, \end{align*} with normal convergence on compact subsets of $B_n$. Fix $\beta\in\mathbb{N}_0^n$. For $0<r<1$, [uniform convergence](/page/Uniform%20Convergence) on the sphere $rS^{2n-1}:=\{r\xi:\xi\in S^{2n-1}\}$ and the rotational orthogonality on $S^{2n-1}$ give \begin{align*} \int_{S^{2n-1}} f(r\xi)\overline{\xi^\beta}\,d\sigma(\xi) = a_\beta r^{|\beta|} \int_{S^{2n-1}}|\xi^\beta|^2\,d\sigma(\xi). \end{align*} Using polar coordinates $z=r\xi$ in the full ball, with $d\mathcal{L}^{2n}(z)=r^{2n-1}\,d\sigma(\xi)\,d\mathcal{L}^1(r)$, orthogonality of $f$ to $z^\beta$ on $B_n$ gives \begin{align*} 0 &= \int_{B_n}f(z)\overline{z^\beta}\,d\mathcal{L}^{2n}(z)\\ &= \int_0^1 r^{|\beta|+2n-1} \left( \int_{S^{2n-1}} f(r\xi)\overline{\xi^\beta}\,d\sigma(\xi) \right)\,d\mathcal{L}^1(r)\\ &= a_\beta \left(\int_{S^{2n-1}}|\xi^\beta|^2\,d\sigma(\xi)\right) \int_0^1 r^{2|\beta|+2n-1}\,d\mathcal{L}^1(r). \end{align*} Both factors multiplying $a_\beta$ are positive, so $a_\beta=0$ for every $\beta\in\mathbb{N}_0^n$. Thus $f=0$, and $(e_\alpha)_{\alpha\in\mathbb{N}_0^n}$ is an orthonormal basis. For each fixed $w\in B_n$, the evaluation map $E_w:A^2(B_n)\to\mathbb{C}$, $E_w(f):=f(w)$, is represented by $K_{B_n}(\cdot,w)$. Expanding this representing vector in the orthonormal basis and using the reproducing identity gives \begin{align*} K_{B_n}(\cdot,w) = \sum_{\alpha\in\mathbb{N}_0^n} K_{B_n}(e_\alpha,w)e_\alpha = \sum_{\alpha\in\mathbb{N}_0^n} \overline{e_\alpha(w)}e_\alpha, \end{align*} with convergence in $A^2(B_n)$ and locally [uniform convergence](/page/Uniform%20Convergence) after evaluation by the bounded point-evaluation functionals. Therefore \begin{align*} K_{B_n}(z,w) &= \sum_{\alpha\in\mathbb{N}_0^n} e_\alpha(z)\overline{e_\alpha(w)}\\ &= \frac{1}{\pi^n} \sum_{\alpha\in\mathbb{N}_0^n} \frac{(n+|\alpha|)!}{\alpha!} z^\alpha\overline{w^\alpha}. \end{align*} Group the sum by $m=|\alpha|$. For $z,w\in B_n$, $|\langle z,w\rangle|<1$, and the multinomial expansion gives \begin{align*} \sum_{|\alpha|=m}\frac{m!}{\alpha!}z^\alpha\overline{w^\alpha} = \left(\sum_{j=1}^n z_j\overline{w_j}\right)^m = \langle z,w\rangle^m. \end{align*} Thus \begin{align*} K_{B_n}(z,w) &= \frac{1}{\pi^n} \sum_{m=0}^\infty \frac{(n+m)!}{m!}\langle z,w\rangle^m\\ &= \frac{n!}{\pi^n} \sum_{m=0}^\infty \binom{n+m}{n}\langle z,w\rangle^m\\ &= \frac{n!}{\pi^n} \frac{1}{(1-\langle z,w\rangle)^{n+1}}. \end{align*} [guided] We now turn the norm computation into the kernel. Define \begin{align*} e_\alpha:B_n&\to\mathbb{C}\\ z&\mapsto \left(\frac{(n+|\alpha|)!}{\pi^n\alpha!}\right)^{1/2}z^\alpha. \end{align*} The previous step gives \begin{align*} \|z^\alpha\|_{A^2(B_n)}^2=\frac{\pi^n\alpha!}{(n+|\alpha|)!}, \end{align*} so each $e_\alpha$ has norm $1$. The rotational argument from the first step gives orthogonality for distinct multi-indices, hence $(e_\alpha)_{\alpha\in\mathbb{N}_0^n}$ is an orthonormal sequence. We also need completeness. Suppose $f\in A^2(B_n)$ is orthogonal to every monomial. Since $f$ is holomorphic, it has a Taylor expansion at $0$, \begin{align*} f(z)=\sum_{\alpha\in\mathbb{N}_0^n} a_\alpha z^\alpha, \end{align*} which converges normally on compact subsets of $B_n$. Fix a multi-index $\beta\in\mathbb{N}_0^n$. We isolate $a_\beta$ by angular integration on spheres, not by restricting the $B_n$ inner product to a smaller ball. For $0<r<1$, define the sphere of radius $r$ by \begin{align*} rS^{2n-1}:=\{r\xi:\xi\in S^{2n-1}\}. \end{align*} The Taylor series converges uniformly on $rS^{2n-1}$, so termwise integration over $S^{2n-1}$ is valid. The same rotation argument used for monomial orthogonality gives \begin{align*} \int_{S^{2n-1}}\xi^\alpha\overline{\xi^\beta}\,d\sigma(\xi)=0 \end{align*} whenever $\alpha\neq\beta$. Hence \begin{align*} \int_{S^{2n-1}} f(r\xi)\overline{\xi^\beta}\,d\sigma(\xi) = a_\beta r^{|\beta|} \int_{S^{2n-1}}|\xi^\beta|^2\,d\sigma(\xi). \end{align*} Now use the original orthogonality on the whole ball. In polar coordinates $z=r\xi$, the domain is $0<r<1$, $\xi\in S^{2n-1}$, and Lebesgue measure transforms as \begin{align*} d\mathcal{L}^{2n}(z)=r^{2n-1}\,d\sigma(\xi)\,d\mathcal{L}^1(r). \end{align*} Since $f$ is orthogonal to $z^\beta$ on $B_n$, \begin{align*} 0 &= \int_{B_n}f(z)\overline{z^\beta}\,d\mathcal{L}^{2n}(z)\\ &= \int_0^1 r^{|\beta|+2n-1} \left( \int_{S^{2n-1}} f(r\xi)\overline{\xi^\beta}\,d\sigma(\xi) \right)\,d\mathcal{L}^1(r)\\ &= a_\beta \left(\int_{S^{2n-1}}|\xi^\beta|^2\,d\sigma(\xi)\right) \int_0^1 r^{2|\beta|+2n-1}\,d\mathcal{L}^1(r). \end{align*} The spherical integral is positive because $|\xi^\beta|^2$ is continuous and not identically zero on $S^{2n-1}$, and the radial integral is positive. Therefore $a_\beta=0$. As $\beta$ was arbitrary, every Taylor coefficient of $f$ vanishes, so $f=0$. This proves that the normalized monomials form an orthonormal basis of $A^2(B_n)$. Now we justify the kernel expansion from this basis. For each fixed $w\in B_n$, the evaluation functional $E_w:A^2(B_n)\to\mathbb{C}$, $E_w(f):=f(w)$, is represented by $K_{B_n}(\cdot,w)$ under the inner product. Expanding this representing vector in the orthonormal basis gives \begin{align*} K_{B_n}(\cdot,w) = \sum_{\alpha\in\mathbb{N}_0^n}(K_{B_n}(\cdot,w),e_\alpha)_{A^2(B_n)}e_\alpha. \end{align*} Because the inner product is conjugate-linear in the second variable and the reproducing identity gives $e_\alpha(w)=(e_\alpha,K_{B_n}(\cdot,w))_{A^2(B_n)}$, we have \begin{align*} (K_{B_n}(\cdot,w),e_\alpha)_{A^2(B_n)}=\overline{e_\alpha(w)}. \end{align*} Thus the series converges to $K_{B_n}(\cdot,w)$ in $A^2(B_n)$. Evaluating at $z$ by the bounded point-evaluation functional gives \begin{align*} K_{B_n}(z,w) = \sum_{\alpha\in\mathbb{N}_0^n} e_\alpha(z)\overline{e_\alpha(w)}. \end{align*} Substituting the definition of $e_\alpha$ gives \begin{align*} K_{B_n}(z,w) &= \frac{1}{\pi^n} \sum_{\alpha\in\mathbb{N}_0^n} \frac{(n+|\alpha|)!}{\alpha!} z^\alpha\overline{w^\alpha}. \end{align*} Now group terms by the total degree $m=|\alpha|$: \begin{align*} K_{B_n}(z,w) &= \frac{1}{\pi^n} \sum_{m=0}^\infty \sum_{|\alpha|=m} \frac{(n+m)!}{\alpha!} z^\alpha\overline{w^\alpha}. \end{align*} For fixed $m$, factor \begin{align*} \frac{(n+m)!}{\alpha!} = \frac{(n+m)!}{m!}\frac{m!}{\alpha!}. \end{align*} The multinomial expansion gives \begin{align*} \sum_{|\alpha|=m}\frac{m!}{\alpha!}z^\alpha\overline{w^\alpha} = \left(\sum_{j=1}^n z_j\overline{w_j}\right)^m = \langle z,w\rangle^m. \end{align*} Since $z,w\in B_n$, the [Cauchy-Schwarz inequality](/theorems/432) in $\mathbb{C}^n$ gives \begin{align*} |\langle z,w\rangle|\le |z|\,|w|<1, \end{align*} so the following [power series](/page/Power%20Series) converges absolutely: \begin{align*} K_{B_n}(z,w) &= \frac{1}{\pi^n} \sum_{m=0}^\infty \frac{(n+m)!}{m!}\langle z,w\rangle^m\\ &= \frac{n!}{\pi^n} \sum_{m=0}^\infty \binom{n+m}{n}\langle z,w\rangle^m. \end{align*} The binomial series identity \begin{align*} \frac{1}{(1-t)^{n+1}} = \sum_{m=0}^\infty \binom{n+m}{n}t^m, \qquad |t|<1, \end{align*} applied with $t=\langle z,w\rangle$, gives \begin{align*} K_{B_n}(z,w) = \frac{n!}{\pi^n} \frac{1}{(1-\langle z,w\rangle)^{n+1}}. \end{align*} [/guided] [/step] [step:Compute the exact $A^2(\Delta^n)$ norm of each monomial] For $\alpha\in\mathbb{N}_0^n$, Tonelli's theorem applies to the nonnegative integrand $|z^\alpha|^2$ on $\Delta^n$, giving \begin{align*} \int_{\Delta^n}|z^\alpha|^2\,d\mathcal{L}^{2n}(z) &= \prod_{j=1}^n \int_{\Delta}|z_j|^{2\alpha_j}\,d\mathcal{L}^2(z_j). \end{align*} Using one-dimensional polar coordinates $z_j=re^{i\theta}$ on $\Delta$, with $d\mathcal{L}^2(z_j)=r\,d\mathcal{L}^1(r)\,d\mathcal{L}^1(\theta)$, we obtain \begin{align*} \int_{\Delta}|z_j|^{2\alpha_j}\,d\mathcal{L}^2(z_j) &= 2\pi\int_0^1 r^{2\alpha_j+1}\,d\mathcal{L}^1(r)\\ &= \frac{\pi}{\alpha_j+1}. \end{align*} Therefore \begin{align*} \left\|z^\alpha\right\|_{A^2(\Delta^n)}^2 = \prod_{j=1}^n\frac{\pi}{\alpha_j+1} = \frac{\pi^n}{\prod_{j=1}^n(\alpha_j+1)}. \end{align*} [guided] For the polydisc, the product structure makes the norm computation coordinatewise. Fix $\alpha\in\mathbb{N}_0^n$. The function $z\mapsto |z^\alpha|^2$ is nonnegative and measurable on $\Delta^n$, so Tonelli's theorem applies to the product measure $\mathcal{L}^{2n}=\mathcal{L}^2\otimes\cdots\otimes\mathcal{L}^2$ on $\mathbb{C}^n$. Since \begin{align*} |z^\alpha|^2 = \prod_{j=1}^n |z_j|^{2\alpha_j}, \end{align*} we obtain \begin{align*} \int_{\Delta^n}|z^\alpha|^2\,d\mathcal{L}^{2n}(z) &= \prod_{j=1}^n \int_{\Delta}|z_j|^{2\alpha_j}\,d\mathcal{L}^2(z_j). \end{align*} Now compute the one-dimensional factor. In the disc $\Delta$, use polar coordinates $z_j=re^{i\theta}$ with $0<r<1$ and $0\le \theta<2\pi$. Under this substitution, planar Lebesgue measure transforms as \begin{align*} d\mathcal{L}^2(z_j)=r\,d\mathcal{L}^1(r)\,d\mathcal{L}^1(\theta). \end{align*} Therefore \begin{align*} \int_{\Delta}|z_j|^{2\alpha_j}\,d\mathcal{L}^2(z_j) &= \int_0^{2\pi}\int_0^1 r^{2\alpha_j}r\,d\mathcal{L}^1(r)\,d\mathcal{L}^1(\theta)\\ &= 2\pi\int_0^1 r^{2\alpha_j+1}\,d\mathcal{L}^1(r)\\ &= \frac{\pi}{\alpha_j+1}. \end{align*} Multiplying the $n$ coordinate factors gives \begin{align*} \left\|z^\alpha\right\|_{A^2(\Delta^n)}^2 = \prod_{j=1}^n\frac{\pi}{\alpha_j+1} = \frac{\pi^n}{\prod_{j=1}^n(\alpha_j+1)}. \end{align*} [/guided] [/step] [step:Sum the normalized monomial expansion for the polydisc] Define, for $\alpha\in\mathbb{N}_0^n$, \begin{align*} u_\alpha:\Delta^n&\to\mathbb{C}\\ z&\mapsto \left(\frac{\prod_{j=1}^n(\alpha_j+1)}{\pi^n}\right)^{1/2}z^\alpha. \end{align*} By the orthogonality and norm computation above, $(u_\alpha)_{\alpha\in\mathbb{N}_0^n}$ is an orthonormal sequence. To prove completeness, let $f\in A^2(\Delta^n)$ be orthogonal to every monomial, and write \begin{align*} f(z)=\sum_{\alpha\in\mathbb{N}_0^n}a_\alpha z^\alpha \end{align*} with normal convergence on compact subsets of $\Delta^n$. Fix $\beta\in\mathbb{N}_0^n$. For $\rho=(\rho_1,\dots,\rho_n)\in(0,1)^n$, define the coordinate torus \begin{align*} T_\rho:=\{(\rho_1e^{i\theta_1},\dots,\rho_ne^{i\theta_n}):0\leq\theta_j<2\pi\text{ for }1\leq j\leq n\}. \end{align*} [Uniform convergence](/page/Uniform%20Convergence) on $T_\rho$ and angular orthogonality give \begin{align*} \int_{[0,2\pi)^n} f(\rho_1e^{i\theta_1},\dots,\rho_ne^{i\theta_n}) \prod_{j=1}^n \rho_j^{\beta_j}e^{-i\beta_j\theta_j} \,d\mathcal{L}^n(\theta) = (2\pi)^n a_\beta\prod_{j=1}^n\rho_j^{2\beta_j}. \end{align*} Using polydisc polar coordinates $z_j=\rho_je^{i\theta_j}$ on the full domain, with \begin{align*} d\mathcal{L}^{2n}(z)=\prod_{j=1}^n\rho_j\,d\mathcal{L}^n(\rho)\,d\mathcal{L}^n(\theta), \end{align*} orthogonality of $f$ to $z^\beta$ on $\Delta^n$ gives \begin{align*} 0 &= \int_{\Delta^n}f(z)\overline{z^\beta}\,d\mathcal{L}^{2n}(z)\\ &= (2\pi)^n a_\beta \prod_{j=1}^n\int_0^1 \rho_j^{2\beta_j+1}\,d\mathcal{L}^1(\rho_j). \end{align*} The product is positive, so $a_\beta=0$. Thus every Taylor coefficient vanishes, and $(u_\alpha)_{\alpha\in\mathbb{N}_0^n}$ is an orthonormal basis of $A^2(\Delta^n)$. For fixed $w\in\Delta^n$, the boundedness of point evaluations proved in the first step allows us to expand the representing vector $K_{\Delta^n}(\cdot,w)$ in this orthonormal basis and then evaluate at $z$, giving \begin{align*} K_{\Delta^n}(z,w) &= \sum_{\alpha\in\mathbb{N}_0^n}u_\alpha(z)\overline{u_\alpha(w)}\\ &= \frac{1}{\pi^n} \sum_{\alpha\in\mathbb{N}_0^n} \left(\prod_{j=1}^n(\alpha_j+1)\right) \prod_{j=1}^n(z_j\overline{w_j})^{\alpha_j}. \end{align*} Since $z,w\in\Delta^n$, each $|z_j\overline{w_j}|<1$. Absolute convergence allows the multi-index series to factor: \begin{align*} K_{\Delta^n}(z,w) &= \frac{1}{\pi^n} \prod_{j=1}^n \left( \sum_{m=0}^\infty (m+1)(z_j\overline{w_j})^m \right). \end{align*} For $|t|<1$, \begin{align*} \sum_{m=0}^\infty (m+1)t^m=\frac{1}{(1-t)^2}. \end{align*} Applying this with $t=z_j\overline{w_j}$ gives \begin{align*} K_{\Delta^n}(z,w) = \frac{1}{\pi^n} \prod_{j=1}^n \frac{1}{(1-z_j\overline{w_j})^2}. \end{align*} This is the asserted Bergman kernel of the unit polydisc. [guided] Define the normalized monomial \begin{align*} u_\alpha:\Delta^n&\to\mathbb{C}\\ z&\mapsto \left(\frac{\prod_{j=1}^n(\alpha_j+1)}{\pi^n}\right)^{1/2}z^\alpha. \end{align*} The orthogonality from the rotation argument and the norm computation from the preceding step show that $(u_\alpha)_{\alpha\in\mathbb{N}_0^n}$ is an orthonormal sequence in $A^2(\Delta^n)$. We next verify completeness. Let $f\in A^2(\Delta^n)$ be orthogonal to every monomial, and write the Taylor expansion of $f$ at $0$ as \begin{align*} f(z)=\sum_{\alpha\in\mathbb{N}_0^n}a_\alpha z^\alpha, \end{align*} with normal convergence on compact subsets of $\Delta^n$. Fix $\beta\in\mathbb{N}_0^n$. The coefficient $a_\beta$ is isolated by angular integration on coordinate tori. For $\rho=(\rho_1,\dots,\rho_n)\in(0,1)^n$, define \begin{align*} T_\rho:=\{(\rho_1e^{i\theta_1},\dots,\rho_ne^{i\theta_n}):0\leq\theta_j<2\pi\text{ for }1\leq j\leq n\}. \end{align*} The Taylor series converges uniformly on $T_\rho$, so termwise angular integration is valid. Since \begin{align*} \int_0^{2\pi}e^{i(\alpha_j-\beta_j)\theta_j}\,d\mathcal{L}^1(\theta_j) = 0 \end{align*} when $\alpha_j\neq\beta_j$, all Taylor terms except $\alpha=\beta$ vanish after integrating over $[0,2\pi)^n$. Thus \begin{align*} \int_{[0,2\pi)^n} f(\rho_1e^{i\theta_1},\dots,\rho_ne^{i\theta_n}) \prod_{j=1}^n \rho_j^{\beta_j}e^{-i\beta_j\theta_j} \,d\mathcal{L}^n(\theta) = (2\pi)^n a_\beta\prod_{j=1}^n\rho_j^{2\beta_j}. \end{align*} Now use the original orthogonality on the whole polydisc. In coordinate polar variables $z_j=\rho_je^{i\theta_j}$, the domain is $0<\rho_j<1$ and $0\leq\theta_j<2\pi$, and the product Lebesgue measure transforms as \begin{align*} d\mathcal{L}^{2n}(z)=\prod_{j=1}^n\rho_j\,d\mathcal{L}^n(\rho)\,d\mathcal{L}^n(\theta). \end{align*} Therefore \begin{align*} 0 &= \int_{\Delta^n}f(z)\overline{z^\beta}\,d\mathcal{L}^{2n}(z)\\ &= (2\pi)^n a_\beta \prod_{j=1}^n\int_0^1 \rho_j^{2\beta_j+1}\,d\mathcal{L}^1(\rho_j). \end{align*} Every radial factor is positive, so $a_\beta=0$. Hence every Taylor coefficient of $f$ vanishes, and the normalized monomials form an orthonormal basis. For fixed $w\in\Delta^n$, the Bergman kernel vector $K_{\Delta^n}(\cdot,w)$ represents the evaluation functional $E_w(f):=f(w)$. Expanding this vector in the orthonormal basis gives \begin{align*} K_{\Delta^n}(z,w) &= \sum_{\alpha\in\mathbb{N}_0^n}u_\alpha(z)\overline{u_\alpha(w)}\\ &= \frac{1}{\pi^n} \sum_{\alpha\in\mathbb{N}_0^n} \left(\prod_{j=1}^n(\alpha_j+1)\right) \prod_{j=1}^n(z_j\overline{w_j})^{\alpha_j}, \end{align*} where the series converges in $A^2(\Delta^n)$ and after point evaluation. Now use the product structure of the multi-index sum. Since $z,w\in\Delta^n$, each scalar $z_j\overline{w_j}$ has modulus less than $1$. Therefore the series is absolutely convergent and factors coordinate-by-coordinate: \begin{align*} K_{\Delta^n}(z,w) &= \frac{1}{\pi^n} \prod_{j=1}^n \left( \sum_{m=0}^\infty (m+1)(z_j\overline{w_j})^m \right). \end{align*} For $|t|<1$, differentiating the geometric series $\sum_{m=0}^\infty t^m=(1-t)^{-1}$ gives \begin{align*} \sum_{m=0}^\infty (m+1)t^m=\frac{1}{(1-t)^2}. \end{align*} Applying this identity with $t=z_j\overline{w_j}$ for each coordinate yields \begin{align*} K_{\Delta^n}(z,w) = \frac{1}{\pi^n} \prod_{j=1}^n \frac{1}{(1-z_j\overline{w_j})^2}. \end{align*} This is exactly the claimed formula for the Bergman kernel of $\Delta^n$. [/guided] [/step]