I figured I can somehow use Liouville's theorem. I don't really know what the method is for solving or where to begin.
Suppose that $f: \mathbb{C} \to \mathbb{C}$ is an entire function. Let $g(z) = f(\frac{1}{z})$. Prove that if $g$ has a pole at $0$ then there is a $z_{0}$ in $\mathbb{C}$ such that $f(z_{0}) = 0$. Can assume $g$ is holomorphic on $\mathbb{C} \setminus \{0\}$.
On
You will need the minimum modulus principle: If the modulus of a holomorphic function on a connected domain has a minimum, then either the function is constant or the minimum is $0$. In this case, you should show that:
a) $f$ is not constant.
b) There is a suitably large open disc $U_R(0)$ on which $\vert f\vert$ has a minimum.
For a), show that if $f$ were constant, then $g$ would have a removable singularity.
For b), take a compact disc $\overline{U_r(0)}$ and show that $\vert f\vert$ has a minimum $m$ on said disc. Then show that there exists another, larger compact disc $\overline{U_R(0)}$ where $\vert f(z)\vert>m$ for all $z\in\partial U_R(0)$. Conclude that $\vert f\vert$ has a minimum on the interior of $U_R(0)$. Then apply the minimum modulus principle.
Since $f$ is entire, there is a power series $\sum_{n=0}^\infty a_nz^n$ centered at $0$ such that$$(\forall z\in\Bbb C):f(z)=\sum_{n=0}^\infty a_nz^n$$and therefore$$(\forall z\in\Bbb C\setminus\{0\}):g(z)=a_0+\frac{a_1}z+\frac{a_2}{z^2}+\cdots$$So, since $g$ has a pole at $0$, there is some $N\in\Bbb N$ such that $n>N\implies a_n=0$. So, $f$ is a polynomial function and therefore you can use the Fundamental Theorem of Algebra to prove that it has a zero.