If $f: \mathbb{C} \rightarrow \mathbb{C}$ is analytic and $\lim_{z \to \infty} f(z) = \infty$ show that $f$ is a polynomial I'm learning about complex analysis and need some help with this problem: 

If $f: \mathbb{C} \rightarrow \mathbb{C}$ is analytic and $\lim_{z \to \infty} f(z) = \infty$ show that $f$ is a polynomial (hint: consider the function $g(z) = f(1/z)$).

Recall that poles are points where evaluating the function would entail dividing by zero. Therefore, since $\lim_{z \to \infty} f(z) = \infty$ this means that $\infty$ is a pole of $f$. How do I continue from here and make use of the hint?

I should mention that this problem has already been asked by other members but I could not find any solution using the given hint. 
 A: Suppose $f$ has Taylor series
$$
f(z)=\sum_{n=0}^{\infty}a_nz^n \quad\text{and }\quad g(z) =f(1/z) =\sum_{n=0}^{\infty}\frac{a_n}{z^n}
$$
If $f$ is not polynomial, then $0$ is an essential singularity of $g$ ($\infty$ is an essential singularity of $f$). By Casorati-Weierstrass theorem, for any $A\in \Bbb{C}$, there is a sequence $z_n\to0$ such that $\lim_{n\to\infty}g(z_n)=A$, i.e. there is $z_n'=1/z_n\to\infty$ such that $\lim_{n\to\infty}f(z_n')=A$, contradicting $\lim_{n\to\infty}f(z)=\infty$.
A: Without Laurent series and assuming $\;f(z)\;$ isn't zero (because then it is trivially true).
By the given information there exists $\;M\in\Bbb R^+\;$ such that $\;|f(z)|>1\;\;\;\forall\,z\in\Bbb C\;\;\text{with}\;\;|z|>M\;$ .
It must be that $\;f(z)\;$ has a finite number of zeros $\;z_1,...,z_n\;$, otherwise its set of zeros, which is in $\;C_M:=\{\,z\in\Bbb C\;;\;|z|\le M\}\;$, has an accumulation point by Bolzano-Weierstrass, and thus from the identity theorem this would mean $\;f(z)=0\;$ .
From here that $\;g(z):=\frac{f(z)}{\prod\limits_{k=1}^n(z-z_k)}\;$ is analytic and non-zero, and thus also $\;h(z)=\frac1{g(z)}\;$ is, and we have for $\;z\in\Bbb C\setminus C_M\;$:
$$|h(z)|=\frac{|z^n+A|}{|f(z)|}\le|z^n|+A\implies h(z)\;\;\text{is a polynomial without roots}\;(**)\;\implies h(z)=K$$
a constant, by the Fundamental Theorem of Algebra, and thus also is $\;g(z)\;$ :
$$\frac{f(z)}{\prod\limits_{k=1}^n(z-z_k)}=g(z)=\frac1{h(z)}=\frac1K\implies f(z)=K(z-z_1)\cdot\ldots\cdot(z-z_n)$$
and we've finished.
If you need a prove of $\;(**)\;$ ask back.
A: Hint: Expand $g(z)$ by Laurent series.
