Let $f:mathbb Crightarrow mathbb C$ be an entire function such that the function $g(z)$ given by $g(z)=f(frac{1}{z})$ has a pole at 0. Prove or disprove $f$ is onto.

I think $f$ is an onto function.

My attempt:

Since $f$ is an entire function $f$ has a power series representation about 0 given by $sum_{m=0}^infty a_m z^m$. Then $g(z)=sum_{m=0}^infty frac {a_m}{z^m}$. Since $g$ has pole at 0 then $g$ is of the form $g(z)=a_0+frac{a_1}{z}+frac{a_2}{z^2}+…+frac{a_n}{z^n}$ for some fixed $nin mathbb N$. Then $f(z)=a_0+a_1z+a_2z^2+….+a_nz^n$. Since $f$ is a polynomial it follows from Fundamental Theorem of Algebra that $f$ is onto.