[proofplan] The proof is an unpacking of the representing-vector definition of the Bergman kernel. First we expand the identity $f(z)=(f,k_z)_{A^2(\Omega)}$ using the Hilbert-space inner product and the definition $K_\Omega(z,w)=\overline{k_z(w)}$. Then we apply the same representing identity to the special functions $k_w$ and $k_z$ to compare $K_\Omega(w,z)$ and $K_\Omega(z,w)$. Finally, setting the two variables equal identifies the diagonal value with the squared norm of $k_z$, which gives nonnegativity. [/proofplan] [step:Expand the representing identity into the integral formula] Fix $f \in A^2(\Omega)$ and $z \in \Omega$. By the defining property of the representing vector $k_z \in A^2(\Omega)$, \begin{align*} f(z)=(f,k_z)_{A^2(\Omega)}. \end{align*} Using the definition of the inner product on $A^2(\Omega)$, this gives \begin{align*} f(z) &=\int_\Omega f(w)\,\overline{k_z(w)}\,d\mathcal{L}^{2n}(w). \end{align*} By the definition of the Bergman kernel, $K_\Omega(z,w)=\overline{k_z(w)}$ for every $w \in \Omega$. Substituting this into the preceding identity yields \begin{align*} f(z)=\int_\Omega K_\Omega(z,w)f(w)\,d\mathcal{L}^{2n}(w). \end{align*} [/step] [step:Compare the kernel values by applying the reproducing identity to kernel vectors] Fix $z,w \in \Omega$. Since $k_w \in A^2(\Omega)$, the representing identity at the point $z$ applies to the function $k_w:\Omega \to \mathbb{C}$ and gives \begin{align*} k_w(z)=(k_w,k_z)_{A^2(\Omega)}. \end{align*} Taking complex conjugates and using conjugate symmetry of the Hilbert inner product, \begin{align*} \overline{k_w(z)} &=\overline{(k_w,k_z)_{A^2(\Omega)}} \\ &=(k_z,k_w)_{A^2(\Omega)}. \end{align*} Again applying the representing identity, now at the point $w$ to the function $k_z:\Omega \to \mathbb{C}$, gives \begin{align*} k_z(w)=(k_z,k_w)_{A^2(\Omega)}. \end{align*} Therefore \begin{align*} \overline{k_w(z)}=k_z(w). \end{align*} Using the definition of $K_\Omega$ twice, \begin{align*} K_\Omega(w,z) &=\overline{k_w(z)} \\ &=k_z(w) \\ &=\overline{\overline{k_z(w)}} \\ &=\overline{K_\Omega(z,w)}. \end{align*} [/step] [step:Identify the diagonal value with the squared norm of the kernel vector] Fix $z \in \Omega$. Applying the representing identity at $z$ to the function $k_z:\Omega \to \mathbb{C}$ gives \begin{align*} k_z(z)=(k_z,k_z)_{A^2(\Omega)}=\|k_z\|_{A^2(\Omega)}^2. \end{align*} By the definition of $K_\Omega$, \begin{align*} K_\Omega(z,z)=\overline{k_z(z)}. \end{align*} The identity just proved gives $k_z(z)=\|k_z\|_{A^2(\Omega)}^2$, which is real and nonnegative. Hence \begin{align*} K_\Omega(z,z)=\overline{k_z(z)}=\|k_z\|_{A^2(\Omega)}^2\ge 0. \end{align*} This proves both the reproducing formula and the stated symmetry and diagonal identities. [/step]