= Solution
Every entire <holomorphic function> has a <Taylor series> $f(z)=\sum_{n\geq0}a_nz^n$, uniformly convergent on each circle. Angular <orthogonality> gives
$$
\frac1{2\pi}\int_0^{2\pi}|f(re^{i\theta})|^2\,d\theta=\sum_{n\geq0}|a_n|^2r^{2n}.
$$
Integrate in polar coordinates and use <Tonelli theorem> for the nonnegative sum:
$$
\|f\|_{\mathcal F}^2
=2\sum_{n\geq0}|a_n|^2\int_0^\infty r^{2n+1}e^{-r^2}\,dr
=\sum_{n\geq0}n!|a_n|^2.
$$
The last integral equals $n!/2$ by substituting $u=r^2$ and repeated <integration by parts>.
Conversely, for any $(b_n)\in\ell^2$, set $a_n=b_n/\sqrt{n!}$. On $|z|\leq R$, the <Cauchy-Schwarz inequality> gives
$$
\sum_{n\geq0}|a_nz^n|
\leq\left(\sum_n|b_n|^2\right)^{1/2}\left(\sum_n\frac{R^{2n}}{n!}\right)^{1/2}
=\|(b_n)\|_{\ell^2}e^{R^2/2}.
$$
Applied to the tails, the same bound proves uniform convergence on every disk. The series therefore defines an entire <holomorphic function>, and the norm identity puts it in the <Bargmann-Fock space> $\mathcal F$. The coefficient map $f\mapsto(\sqrt{n!}a_n)$ is consequently an onto linear isometry $\mathcal F\to\ell^2$; polarization also identifies the inner products. Completeness of $\ell^2$ proves that $\mathcal F$ is a <Hilbert space>. Its standard coordinate <orthonormal basis> corresponds to
$$
\boxed{e_n(z)=\frac{z^n}{\sqrt{n!}},\quad n=0,1,\ldots.}
$$
The norm identity shows both their unit norms and completeness; it also gives $|f(z)|\leq e^{|z|^2/2}\|f\|_{\mathcal F}$.
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