Need help to prove power series I try to prove the following series:
$\frac{1}{(1-z)^{2}} = \sum_{n=0}^{\infty} (n+1)z^{n} \quad z \in \mathbb{C}, \quad |z| &lt; 1$

What I have tries so far:

$\text{I tries to use }\frac{1}{1-z} = \sum_{n=0}^{\infty} z^{n}\quad |z| &lt; 1 \text{ and } \frac{1}{1-2z + z^2} = \frac{1}{1-(2z-z^2)}$

$\Rightarrow \sum_{n=0}^{\infty} (2z-z^2)^{n} = \sum_{n=0}^{\infty} z^n(2-z)^n$

but I can't go anyway
You can see it in two different ways. As Michael Galuza has suggest you have: 
$\frac{1}{(1-z)^2}=\big(\frac{1}{1-z}\big)'$ then 

$$\Big(\sum_{\mathbb{N}}z^n\Big)'=\sum_{\mathbb{N}}(n+1)z^n.$$

Otherwise you can observe that:

$\frac{1}{(1-z)^2}=\big(\frac{1}{1-z}\big)^2$ so 

$$\Big(\sum_{\mathbb{N}}z^n\Big)\Big(\sum_{\mathbb{N}}z^n\Big)=\sum_{n\in\mathbb{N}}\sum_{i=0}^n(z^i z^{n-i})=\sum_{n\in\mathbb{N}}(n+1)z^n$$

because $\sum_{i=0}^n(z^i z^{n-i})=(n+1)z^n$.
I think I almost found the solution:

$\frac{1}{(1-z)} = \sum_{n=0}^{\infty} z^n$

$\Rightarrow (\frac{1}{(1-z)})' = (\sum_{n=0}^{\infty} z^n)'$

$(1-z)^{-2} = \sum_{n=1}^{\infty} nz^{n-1} = \sum_{n=0}^{\infty}(n+1)z^n$