Write Taylor Expansion Of $\frac{1}{(1-z)^2}$ 
Write Taylor Expansion Of $\frac{1}{(1-z)^2}$  around $z_0=0$

I need to "get rid" of the 2 power and I can use $\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$ but how can I do that?

Since$$\frac1{1-z}=1+z+z^2+z^3+\cdots\text,$$then $\frac1{(1-z)^2}$ is the Cauchy product of the geometric series by itself. So,\begin{align*}\frac1{(1-z)^2}&=(1+z+z^2+z^3+\cdots)(1+z+z^2+z^3+\cdots)\\&=1+2z+3z^2+4z^3+\cdots\end{align*}
