[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]
