Prove $f$ entire and with a pole at infinity to be polynomial The question has been asked before, but I can't seem to wrap my head around it. I guess I'm missing something.
There is a certain part where I'm experiencing problems, and I can't seem to find an answer to that specific part in the other question(s).

If $f$ is entire and $\lim_{z\to \infty} f(z)= \infty$ then $f$ is  polynomial.

Proof:
Since $f$ is entire it can be expanded as a Taylor-series.
$$f(z) =\sum_{n=0}^{+\infty} a_n z^n$$
To prove $f$ would be polynomial I would have to prove $(\exists N)(\forall n > N)(a_n = 0)$
Define:
$$g(z) = f\left( \frac{1}{z}\right) = \sum_{n=0}^{+\infty} a_n\left(\frac{1}{z}\right)^n$$
Since $0$ is a pole (see below) of $g(z)$ the expansion above would be it's corresponding Laurent-expansion. And by definition of a pole $(\exists N)(\forall n >N)(a_n = 0)$.
Q.E.D.
However
Why is $0$ a pole of $g(z)$, why does $\lim_{z \to \infty} f(z) = \infty$ imply $0$ to be a pole of $g(z)$? I know $\lim_{z \to \infty} f(z) = \infty$ implies $\infty$ to be a pole of $f(z)$, but then...?
 A: Well, $g(0)=f\left(\frac10\right)=f(\infty)=\infty$.
A: Note that when $z\rightarrow \infty, w=\frac{1}{z}\rightarrow 0$. And since $f(z)\rightarrow \infty$ as $z\rightarrow \infty$, there is no essential singularity. So $g(w)$ must have a pole of finite order at $\infty$. 
A: To show that 0 is a pole for f(1/z), try this:
Define $h(z)=\frac{1}{f(\frac{1}{z})}$, then $h(z)$ is analytic for $f(\frac{1}{z})$ non-zero. And note that $\lim_{z\to 0} |h(z)|=0 \implies lim_{z\to 0}h(z)=0$. 
As the limit exists, we can take $h^*$ to be an analytic extension. Knowing that $h^*(z)$ is not identically zero, gives us that zeros are separated, so we can find a ball of radius $R>0$ such that $h^*(z)$ is analytic on $B_R(0)$.
Thus $h^*$ has a Taylor expansion centered about 0 on that region. As the first coefficient of that is 0 (Looking at the limit), but not all coefficients are non-zero or $h^*$ would be identically zero.
We have that $h^*(z)=z^N*\sum_{n=N}^\infty a_n z^{n-N}=z^N*k(z)$, where k is analytic and doesn't vanish on the disk. 
$$lim_{z\to 0}Z^{N+1}f\left(\frac{1}{z}\right)=lim_{z\to 0} \frac{z}{k(z)}=0$$
Therefore $f(1/z)$ has a pole of order N at 0 $\implies$ f(z) has a pole of order N at infinity. Given this, dietervdf proof follows easily, but proving that the singularity at infinity is a pole and not an essential singularity is, I think, non-trivial.
