Let $f$ be an entire function. Think of it as a covering space of $\mathbb{C}$ (perhaps with isolated punctures) to $\mathbb{C}$ (perhaps with isolated punctures). Suppose we know there is only a finite number of covering transformations, $\{\varphi\}_{i=1}^n$: $$ f(\varphi_i(z)) = f(z).$$ How to show $f$ is a polynomial ?
Partial answer: The function $\frac{f(z)-f(w)}{z-w}$ is entire in both variables. It is only zero when $z=\varphi_i(w)$ for some $i$, so we may write $$\frac{f(z)-f(w)}{z-w} = e^{g(z,w)} \prod_{i=1}^n (z-\varphi_i(w)).$$ Now setting $w=0$, and denoting $w_i = \varphi_i(0)$, we obtain: $$\frac{f(z)-f(0)}{z} = e^{g(z,0)} \prod_{i=1}^n (z-w_i)=$$ $$ e^{g(z,0)} (z^n - (\sum w_i) z^{n-1} + \ldots + (-1)^n \prod w_i). $$ To show this is a polynomial amounts to showing $g(z,0)$ is a constant.
