$f : \Bbb{C} \to \Bbb{C}$ 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!!!
 A: 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 $f(z)$.
