TITLE: entire function with only finitely many zeros
QUESTION [14 upvotes]: I saw the following exercise:
If $f:\mathbb{C}\rightarrow\mathbb{C}$ is an entire, non-constant function with only finitely many zeros, then either $|f(z)|\rightarrow \infty$ for $|z|\rightarrow\infty$ or there is a sequence of points $z_n$ such that $|z_n|\rightarrow\infty$ and $f(z_n)\rightarrow 0$.
I thought a bit about this exercise and of course $f$ has to be unbounded because of Liouville's Theorem. But if I assume, that there is a unbounded sequence $z_n$ for which $f(z_n)\rightarrow \infty$ does not hold, how can I conclude, that there has to be a sequence such that $f(z_n)$ goes to zero? 
Thanks for hints!

REPLY [6 votes]: To add on to what  Makuasi stated an entire function has a pole of order $n$ at $\infty$ iff the function is a polynomial of order $n$.   So if $f$ does not have a pole at $\infty$ $f$ is either bounded or has an essential singularity at $\infty$.   $f$ cannot be bounded by Liouville's Thm.  So $f$ must have an essential singularity at $\infty$.   By Casorati–Weierstrass theorem there is a sequence, $(z_{n})$, such that $|z_{n}| \rightarrow \infty$, and  $f(z_{n})\rightarrow 0$