$f : \Bbb{C} \to \Bbb{C}$ is an entire function s.t. $|f(z)| \to \infty$ as $|z| \to \infty$. Prove $f$ is a polynomial. Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be an entire function such that $|f(z)| \rightarrow  \infty$ as $|z| \rightarrow  \infty$. Prove that $f$ is a polynomial by following the steps below.
(a) Observe that the function $f(1/z)$ defined in $C \setminus\{0\}$ has a pole at the origin. Let $S$ be the
singular part of its Laurent series around $0$. Argue that $g(z) = f(z) − S(1/z)$ approaches
finite limits as $z \rightarrow 0$ and as $|z| \rightarrow  \infty$.
(b) Prove that $g$ extends to a bounded entire function and is therefore constant.
(c) Deduce that $f$ is a polynomial.

I know there are solutions to this but not with these steps. I have been given this as an assignment and really need help to figure out how to write the solution.
For point (a) I wrote 
In order for $|f(z)| \rightarrow  \infty$ as $|z| \rightarrow  \infty$, is equavalent saying $|f(\frac{1}{z})| \rightarrow  \infty$  as $z \rightarrow 0$. So $f(z)$ can be written in Laurent series around the point $0$ such that $\sum_{n=-\infty}^\infty a_nz^n = g(z) + s(\frac{1}{z})$ given $g(z)$ is the power series $\sum_{n=0}^\infty a_nz^n$ and $S$ is the singular part of its Laurent series around $0$.
Hence,
$ g(z)= f(z) - s(\frac{1}{z})$ and $g(z)$ tends to a 
finite limits as $z \rightarrow 0$ and as $|z| \rightarrow  \infty$ given the power series.
For (b) I wrote
Only removable singularities are left in $g(z)$. So
with Riemann's removable singularity theorem, with $p$ being some removable singularity. $g(z)$ can be extended continuously and holomorphically at $p$ and since $g(z)$ is bonded on some disc around $p$ excluding the point $p$ with a radius bigger than $0$. Hence the extended function is bounded as well and by Liouville's theorem, it is a constant function.
And I have no clue how to answer (c), sorry for the long post and thanks in advance!!!

In your solution to part (a), notice that the Laurent series you have written down is for $f(\frac 1z)$, so the correct formula is $f(\frac 1z)=\sum_{n=0}^\infty a_n z^n +S(z)$. So by the change of coordinates $z\mapsto \frac 1z$, we get $f(z)=\sum_{n=0}^\infty a_n\frac 1{z^n} +S(\frac 1z)$. So the $g(z)$ in the question is actually $\sum_{n=0}^\infty a_n \frac 1{z^n}$. As $|z|\to\infty$, this series indeed tends to $a_0$, which is finite. As for $z\to 0$. Now by looking that the Laurent series again, we see that $S(\frac 1z)$ has removable singularity at $z=0$. So $g(z)=f(z)-S(\frac 1z)$ also has removable singularity at $z=0$, in other words the limit as $z\to 0$ exists.
For part (b), holomorphicity of $g(z)=\sum_{n=0}^\infty a_n \frac 1{z^n}$ follows directly from $g$ having removable singularity at $0$. By comparing the power series and Laurent series, we see that the only term appearing in the series must be the constant $a_0$ only.
For part (c), the hint is to look at the Laurent expansion of $f(\frac 1z)$ again, and recover the series for $f(z)$.
