TITLE: If $f$ is an entire function that goes to infinity, $f$ is a polynomial
QUESTION [5 upvotes]: I am asked to show that if $f$ is entire with the property that $\lim_{z\to\infty} f(z)=\infty$, then $f$ must be a polynomial. 
However, I feel as if I am missing something. 
$e^z$ is entire (holomorphic at all finite $z$), and clearly $\lim_{z\to\infty} f(z)=\infty$ holds. However, $e^z$ is not a polynomial. 
This seems to be a disproof via counterexample. Is there something wrong with my reasoning, or with the problem?

REPLY [2 votes]: One important feature about holomorphic functions is the number and position of their zeroes. The function going to $\infty$ implies that outside of some compact disk $D$ centered at the origin, the function is never zero. 
Now to prove that our function is a polynomial, it suffices by analytic continuation to prove that our $f$ has only a finite number of zeroes in this disk $D$. What would go wrong if our function had infinitely many zeroes in $D$? Let $S$ be the set of zeroes of $f$. Since $D$ is compact, $S$ would have a limit point in $D$; but again, by analytic continuation, this would imply that $f$ is the zero function, which is a contradiction.