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?
 A: Take the derivative of that geometric series. Note that in general, we have that
$$\frac{1}{(1-z)^{k+1}} = \frac{1}{k!} \left(\frac{d}{dz}\right)^{k}\frac{1}{1-z}.$$
After a moment's thought, I cannot avoid but give you the following combinatorial proof that the coefficient of $z^n$ in $\frac{1}{(1-z)^{k+1}}$ is equal to 
$$\binom{n+k-1}{k}.$$
To see this, consider the $k+1$-fold product of $\frac{1}{1-z}$ with itself. The coefficient of $z^n$ in such product is the number of ways you can write $n$ as a sum of $k+1$ ordered non-negative integers, $(n_0,\ldots,n_k)$. By adding $1$ to each $n_i$ we can instead consider the problem of writing $n+k+1$ as a sum $k+1$ ordered nonzero integers $(m_0,\ldots,m_k)$. Now this is just tantamount
to lying down $n+k$ balls in a row, and choosing $k$ spots where to divide them,
these spots corresponding to a $k$ subset of $n+k-1$: the place where a bar is found corresponds to the integer in $\{1,\ldots,n+k-1\}$.  
A: 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*}
