I want to prove that if f has an essential singularity at $z_0$ then $f^2$ has an essential singularity at $z_0$ as well.
I know that if f has an essential singularity at $z_0$ then f can be represented as $f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$ where $a_n \neq0$ for an infinite set on which $n<0$.
I also know that if one wants to multiply a series the following is true $\left(\sum_{k=0}^\infty a_k \right) \left(\sum_{k=0}^\infty b_k \right) = \sum_{n=0}^\infty \sum_{k=0}^n a_k b_{n-k}$.
So to prove what I mentioned can I just say that
$(f(z))^2=(\sum_{-\infty}^{\infty}a_n(z-z_0)^n)(\sum_{-\infty}^{\infty}a_n(z-z_0)^n)=\sum_{n=0}^\infty \sum_{k=0}^n a_k a_{n-k}(z-z_0)^n=\sum_{-\infty}^{\infty}c_n(z-z_0)^n.$
Where $c_n=|a_n|^2$ and then note if $a_n \neq0 \forall n<0 \Rightarrow c_n \neq0 \forall n<0$ and so $f^2$ also has an essential singularity at $z_0$ ?