I try to prove the following series:
$\frac{1}{(1-z)^{2}} = \sum\_{n=0}^{\infty} (n+1)z^{n} \quad z \in \mathbb{C}, \quad |z| < 1$

What I have tries so far:

$\text{I tries to use }\frac{1}{1-z} = \sum\_{n=0}^{\infty} z^{n}\quad |z| < 1 \text{ and } \frac{1}{1-2z + z^2} = \frac{1}{1-(2z-z^2)}$

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

but I can't go anyway