Let $f = \sum_{n =0}^{\infty} a_nz^n$ be an entire function such that $\forall w \in \mathbb{C}, f(z) = w$ admits a finite number of solutions.
I want to prove that f is a polynomial.
So far, I know that $f(z) - w$ admits a finite number of roots, by fixing $w$ I wanted to directly show that it forces $f$ to be a polynomial of maximum degree $n$ (with $n$ being the number of roots).
I don't know how to proceed and any help would be welcome.
Note that for any entire function $f$, the function $g(z) = f(1/z)$ has an isolated singularity at $z = 0$. Looking at the power series expansion $f(z) = \sum a_n z^n$ one gets the Laurent series expansion for $g$ at $0$ as $$g(z) = \sum_{-\infty}^0 a_{-n} z^{n}.$$ So it is clear that $f$ is a polynomial (i.e. has finite power series) if and only if $g$ has a pole at $z=0$, instead of an isolated singularity.
Now the claim follows by Picard's great theorem[1]:
If an analytic function $f$ has an essential singularity at a point $w$, then on any punctured neighborhood of $w$ , $f ( z )$ takes on all possible complex values, with at most a single exception, infinitely often.
As $g(z) = f(1/z)$, also $g$ takes on each value finitely many often, especially so in each neighbourhood of $z=0$. Hence $g$ cannot have an essential singularity.
