Prove, if $z=z_0$ is a zero of order $n$ of $f(z)$, then it is a zero of order $2n$ of $(f(z))^2$

Question

Where would you start with this proof. I understand that if $z=z_0$ is a zero of order $n$ of $f(z)$, then it is a zero of order $mn$ of $(f(z))^m$ but do not know how to go about in proving this.

Answer 1

I would start like this: if $z_0$ is a zero of otder $n$ of $f$, then, near $z_0$, $f(z)$ can be written as $(z-z_0)^ng(z)$ where $g$ is analytic and $g(z_0)\neq0$. (I am assuming that $f$ is analytic.)