$\lim\limits_{|z| \to \infty} f(z) = \infty$, show that $f$ is a polynomial. I had an approach to the following problem which now I'm not sure will work:

If $f$ is entire, and $\lim\limits_{z \to \infty} f(z) = \infty$, show that $f$ is a polynomial.

Case 1: there exist $C > 0, N \in \mathbb{N}$ such that $|f(z)| \leq C \cdot |z|^N$ for all $z$.  A standard argument shows that $f$ has to be a polynomial.
Case 2 is where Case 1 does not hold.  Since $f(z)$ goes to infinity, $f$ can have only finitely many zeroes $c_1, ... , c_r$ with multiplicities $m_1, ... , m_r$.  Then I'd like to say that $f$ grows so fast that $$g(z) := \frac{f(z)}{(z-c_1)^{m_1} \cdots (z - c_r)^{m_r}}$$ also goes to infinity as $z \to \infty$.  But it's not clear that this will actually be the case.  I might have to weaken the hypothesis of Case 1 to make this work, but I'm not sure how weak I can make it.  
Assuming I can have $g$ go to infinity, then $g$ will be an entire function with no zeroes, so $\frac{1}{g}$ will be bounded, hence constant.  Thus $f(z) = k(z-c_1)^{m_1} \cdots (z - c_r)^{m_r}$ for some $k \in \mathbb{C}$, as required.
 A: I thought one can make this argument work like this: 
I would start with the argument in case 2. If $g$ is not bounded you are done, as you explained. If $g$ is bounded, say $|g(z)| \le C$, then 
$|f(z)| \le C \prod_{i=1}^r |(z-c_i)^{m_i}|$.
Then remark (maybe you need to give some details there) that $\prod_{i=1}^r |(z-c_i)^{m_i}| \le C' |z|^{(m_1+1) + \dots + (m_r+1)}$ and recall the argument given in case 1.  (That I add the $+1$s is not really needed, but it makes the "remarking" easier.) 
Or if you prefer, modify your case 1 to show that $|f(z)| \le |P(z)|$ for some polynomial $P$ implies that $f$ is  a polynomial.
But I am not sure anymore this line of argument works. Another way to approach this problem, found at Show that this entire function is polynomial. ,  is to consider $z\mapsto f(1/z)$ on $\mathbb{C} \setminus  \{0\}$ instead. Then, at $0$ this function has a pole (and not an essential singularity) as it tends to infinity. Thus the principal part of the Laurent series must be finite, which implies that the power series expansion of $f$ was finite.   
