Holomorphic and Bounded

Question

So, I am trying to prove that if f(z) has a pole at infinity of order n then, its a polynomial. For that assume f(z) has a pole at infinity of order n. So, $f(\frac{1}{z})$ has pole at 0 of order n. So, we can write $f(\frac{1}{n}) = \frac{g(z)}{z^n}$, where g is nonzero and holomorphic around 0. We now have $g(z)=z^n f(\frac{1}{z})$. Consider $\frac{1}{z}$ and with this, we obtain $g(\frac{1}{w}) = \frac{f(w)}{w^n}$. I am trying to show that $g(\frac{1}{w})$ is bounded.

Answer 1

Hint $$g^{(m)}(0)=\frac{n!}{2 \pi i}\int_{|z|=R} \frac{g(z)}{z^{m+1}} dz$$

Make $R \to \infty$ and try to express it as an integral in terms of $f$ around $0$. Prove that $g^{(m)}(0)=0$ for all $m >n$.