TITLE: Show: An entire function $g$ with $\vert g(x) \vert \to \infty$ for $|x| \to \infty$ is a polynomial.
QUESTION [5 upvotes]: This is part of an exercise sheet in complex analysis. It should by solvable by rather elementary methods like the main theorems of complex analysis.
I succeded to show that $g$ has only finitely many zeros by using Bolzano-Weierstraß's theorem. If you divide $g$ by all its zeros you get a holomorphic function without zeros. I now fail to show that this new function is constant.

REPLY [4 votes]: Because it tends to infinity as $|z|$ tends to infinity, it is clear that the function has a pole at infinity. Looking at it on the riemann sphere, this means that it is a meromorphic function, with its only singularity at infinity. But you can prove that all meromorphic functions on the riemann sphere are rational functions, that is $P(z)/Q(z)$ where $P$ and $Q$ are polynomials. But since $Q$ has no zeroes in $\mathbb{C}$ (if it did, the function would not be entire) it must be constant by the fundamental theorem of algebra, so the function is equal to $P(z)$ and is a polynomial.