I am trying to prove the following:
If $f(z)$ is an entire function on $\mathbb{C}$ and $\displaystyle\lim_{|z|\rightarrow \infty} |f(z)|=\infty$ then $f(z)$ is a polynomial.
Let $a\in \mathbb{C}\setminus \{0\}$. Since $f(z)$ is analytic at $a$, it can be expressed as the power series $$f(z)=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (z-a)^n.$$
I have not learned anything about poles in my complex analysis class, so I am expected to solve this without any theorems involving poles. I am trying to work with the function $g(z):=f(1/z)$. This function would have a singularity at zero, but I won't see how to use the function to prove the claim. Is there a way I can show that there exists some $N$ such that $\frac{f^{(n)}(a)}{n!}=0$ for all $n\geq N$?
Also, prior to this I showed that if $f(z)$ is entire and $|f(z)|\leq A+B|z|^n$ for all $z \in \mathbb{C}$ then $f(z)$ is a polynomial. Perhaps I can invoke this result? Any help is greatly appreciated, thank you.