Show that this entire function is polynomial. Let $f$ be an entire function such that $ |f(z)| \to \infty$ as $|z| \to \infty$. Prove that $f$ is a polynomial.
 A: Hint: The set of zeroes of $f$ is bounded and discrete, hence finite and this allows us to find a polynomial $p$ such that $f(z)=p(z)g(z)$ for some entire function $g\colon \mathbb C\to\mathbb C^\times$. What scenarios can happen with $g(z)$ as $z\to\infty$?
A: Consider the holomorphic function on $\mathbb{C}^*$
$$
g(z)=f\left(\frac{1}{z}\right).
$$
So 
$$
\lim_{z\rightarrow 0}\;|g(z)|=+\infty.
$$
Obviously, $0$ is not a removable singularity.
Now by the Casorati-Weierstrass theorem, $0$ is not an essential singularity either. Indeed, take $r>0$ such that $|g(z)|\geq 1$ for all $0<|z|<r$. Then $g(z\;;\;0<|z|<r\})$ is certainly not dense in $\mathbb{C}$.
Edit: As pointed out by @mrf, it is actually easy to see that this implies $0$ is a pole for $g$. It suffices to use Riemann's theorem on removable singularities adequately. Indeed $h(z)=1/g(z)$ is holomorphic and bounded in some $\{z\;;\;0<|z|<r\}$ with limit $0$ at $0$, so $0$ is a removable singularity of $h$ and a zero of the holomorphic extension of $h$. Factor $h(z)=z^nk(z)$ with $k$ holomorphic and $k(0)\neq 0$. The result follows. And this is indeed more elementary than Casorati-Weierstrass.
So keep the Casorati-Weierstrass argument for weaker assumptions such as the ones I mention below.
So $0$ must be a pole. Hence the Laurent series of $g$ at $0$ is of the form
$$
g(z)=\frac{a_n}{z^n}+\ldots+\frac{a_1}{z}+a_0.
$$
Note that there are no positive powers of $z$ because this comes from the power series of $f$.
The result follows clearly by going back to $f$. 
Edit: Note the same proof works if we simply assume that $f$ is entire and there exist $\alpha>0$ and $M>0$ such that
$$
|f(z)|\geq \alpha\qquad \forall |z|\geq M.
$$
Of course, there could be even weaker assumptions like there exists $M>0$ such that $f(\{z\;;\; |z|>M\})$ is not dense in $\mathbb{C}$ and the proof would still work.
