$e^z-P(z)$ has infinitely many zeros

If $P\in\mathbb{C}[z]$ is a non-zero polynomial, prove $f(z):=e^{z}-P(z)$ has infinitely many zeros.

I've made some progress so far, but I still have a step missing. Here is where I'm at:

Suppose $f$ has finitely many zeros, namely $z=a_1,\,...\,,a_n \neq 0$ and possibly $z=0$. Since $e^z$ and $P(z)$ are entire functions with orders of growth $1$ and $0$ respectively, then $f$ has order of growth $1$.

From Hadamard's factorization:

$$f(z)=e^{\alpha z+\beta}z^m\prod_{k=1}^{n}\left(1-\frac{z}{a_k}\right) e^{\left(\frac{z}{a_k}\right)} \,\,\,\,\,\,\,\,\,\,\,\,(*)$$ (where $\alpha$, $\beta$ are complex constants and $m:=$ multiplicity of $z=0$).

Rearranging $(*)$, we get:

$$f(z)=e^{\gamma z+\beta}Q(z)\,\,\,\,\,\,\,\,\,\,\,\,(**)$$

(where $\gamma = \alpha +\sum_{k=1}^{n}\frac{1}{a_k}$ is a constant and $Q(z)=z^m\prod_{k=1}^{n}\left(1-\frac{z}{a_k}\right)$ is a polynomial of degree $m+n$)

Now, for $d:=\text{deg}(P)$, notice $f^{(d+1)}(z)=e^{z}$. Taking the $(d+1)$-th derivative in $(**)$, we get $e^{z}=e^{\gamma z+\beta}R(z)$, where $R$ is a polynomial of degree $m+n$. If $m+n\geq 1$, $R$ has at least one root, while $e^{z}$ has none (absurd), so $m+n=0\Rightarrow m=n=0\Rightarrow f$ has no zeros.

Conclusion: either $f$ has infinitely many zeros or none at all.

Now how can I prove it has at least $1$ zero?

Solution 1:

$P\neq0$ only has finitely many roots, so the function

$$ g:z\mapsto\frac{\exp(1/z)}{P(1/z)} $$

is well defined on a neighbourhood of $0$ (without $0$), has an essential singularity at $0$, and does not vanish. Since it is an analytic function, Great Picard's theorem holds, and $g$ takes all complex values infinitely often except at most one. In particular, $1$ is reached infinitely many times.