Let $H$ be a (separable) Hilbert space, and let $\{e_n\}_{n\in\mathbb{N}}$ be an orthonormal basis. Define the following function \begin{align*} \phi:\mathbb{C}&\to H\\ z&\mapsto \sum_{n\in\mathbb{N}}\frac{z^n}{n!}e_n \end{align*}
Show that $\phi$ is well-defined, i.e. show that $\phi(z)\in H$ for all $z\in\mathbb{C}$.
This part I feel pretty good about. We can estimate the norm by \begin{align*} \| \phi(z) \|^2&=\left\| \sum_{n\in\mathbb{N}}\frac{z^n}{n!}e_n \right\|^2=\sum_{n\in\mathbb{N}}\frac{|z|^{2n}}{n!^2}\leq e^{2|z|}<\infty \end{align*}
Let $S=\text{span}\{\phi(z)\mid z\in\mathbb{C}\}$ be the set of all finite linear combinations of elements in the image of $\phi$. Either show that $S\leq H$ is dense or find a non-zero $v\in S^\perp$.
My intuition is that $S$ is dense, and that I should use the strategy of showing that $S^\perp=\{0\}$. Here's what I tried:
Suppose $v=\sum_{n\in\mathbb{N}}v_ne_n\in S^\perp$. Then $0=\langle v,\phi(0)\rangle=v_0$. To try to show the other coefficients are zero, I noticed that the expansion of $(e^z-1)^N$ gives a way to find a vector $\omega^N\in S$ such that $\langle \omega^N,e_n\rangle=0$ for all $n<N$. Let me write a vector in sequence notation to illustrate what I mean. For $N=1$, we have \begin{align*} e^z-1\Longrightarrow \phi(1)-\phi(0)&=(1,1,\frac{1}{2},\frac{1}{6},\ldots) \\ &-(1,0,0,0,\ldots) \\ &=(0,1,\frac{1}{2},\frac{1}{6},\ldots) \end{align*} For $N=2$, we have $(e^z-1)^2=e^{2z}-2e^{z}+1$ so \begin{align*} \phi(2)-2\phi(1)+\phi(0)&=(1,2,2,\frac{4}{3},\ldots ) \\ &-(2,2,1,\frac{1}{3},\ldots) \\ &+(1,0,0,0,0,\ldots) \\ &=(0,0,1,1,\frac{7}{12},\ldots) \end{align*}
Similarly, $$\omega^N=\sum_{n=0}^N\binom{N}{n}\phi(n)=(0,0,\ldots,0,1,\frac{N}{2},\ldots)$$ Now I'm kinda stuck because I'm not sure how to apply the inductive step to show $v_{N+1}=0$. I'm wondering if this isn't at all the right approach... Another idea would be to use identities like $$\frac{\phi(z)+\phi(-z)}{2}=(1,0,\frac{z^2}{2},0,\frac{z^4}{4!},0,\ldots)$$ Similarly we can use an $m^{th}$ root of unity to find a vector in $S$ such that the coefficient $a_m$ on $e_m$ is only non-zero if $n$ divides $m$. Any tips would be appreciated!