Complex Analysis: Prove that an entire function with $\lim_{z\to\infty}f(z)=\infty$ is surjective. 
If $f$ is an entire function with the property that $|f(z)|\to\infty$
  as $|z|\to\infty$, verify that $f(\mathbb{C})=\mathbb{C}$.

This is a problem from my textbook. And I guess the Rouché Theorem should be applied, but don't know how.
 A: Here's a way to look at it. As $z\to\infty$, $\frac{1}{z}\to 0$. The character of the singularity of a function at infinity is given by replacing $z$ by $\frac{1}{w}$ in its Taylor series and examining the behavior as $w\to 0$. If $f\in \mathcal{H}(\mathbb{C})$ is entire and has $\lim_{z\to\infty} \lvert f(z)\rvert=\infty$, study the Taylor series at $0$
$$ f(z)=\sum_{k=0}^\infty a_k z^k.$$
At infinity, the Laurent series looks like
$$ f(w)=\sum_{k=0}^\infty a_k w^{-k}. $$
If there were infinitely many terms, then $f(w)$ would have an essential singularity at $w$, and $f(z)$ would have an essential singularity at $\infty$. But the $\lvert f(z)\rvert$ would not converge as $z\to\infty$. So, $a_k=0$ for $k\ge M$. This implies that $f(z)$ is a polynomial. 
Can you conclude using Rouché's Theorem?
A: Since $f(z)\to\infty$ as $z\to\infty$, we have $\inf\{\lvert f(z)\rvert:\lvert z\rvert=R\}\to\infty$ as $R\to\infty$.  By Rouché, $f(z)$ and $f(z)-a$ has the same (finite by identity principle) number of roots inside $B_R(0)$ for $\lvert a\rvert<\inf\{\lvert f(z)\rvert:\lvert z\rvert=R\}$.  Use this in two steps, first for $a=f(0)$ and $R$ sufficiently large to conclude $0\in f(\mathbb{C})$, then using a possibly bigger $R$ for other $a$'s.
