If $f$ is an entire function that goes to infinity, $f$ is a polynomial 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? 
 A: I believe you have mis-read or mis-interpreted the key part of the problem. For complex variables,
$$
\lim_{z\to \infty} f(z) = X
$$
means that the limit is $X$ (whatever $X$ is) as $z$ approaches infinity in any direction.
I'm not certain, but I believe you could replace that with
$$
\forall u \in \Bbb C : (|u|=1 \implies \lim_{r\to\infty_{r\in\Bbb R}} f(r u) = X)
$$
that is, the limits, taken as the limit of the real length along a line from the origin, of the function of points on that line, are all the same value $X$ (which in this case is $\infty$.
So you need to show that for any non-polynomial entire function, there is some direction along with the function does not go to infinity. 
A: 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.
A: Let $g(z)=f\left(\frac1z\right)$. Then $\lim_{z\to0}g(z)=\infty$ and therefore $g$ has a pole at $0$, because otherwise $g$ would have an essential singularity at $0$, which is impossible, by the Casorati-Weierstrass theorem. But if $f(z)=a_0+a_1z+a_2z^2+\cdots$, then $g(z)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$ and therefore asserting that $g$ has a pole at $0$ means that $a_n=0$ if $n$ is large enough. So, $f$ is a polynomial function.
