Proving $ \sum_{n=1}^{\infty} nz^{n} = \frac{z}{(1-z)^2}$ for $z \in (-1, 1)$ I do not know where to start, any hints are welcome.

Note that $$\sum_{n=0}^\infty z^n=\frac{1}{1-z} \tag 1$$
for $|z|<1$.  
Differentiating $(1)$ and multiplying by $z$ (this is legitimate since for any $r<1$, $\sum_{n=1}^\infty nz^{n-1}$ converges uniformly for $|z|\le r<1$) yields
$$\begin{align}
z \frac{d}{dz}\sum_{n=0}^\infty z^n&=\sum_{n=0}^\infty nz^n\\\\
&=\sum_{n=1}^\infty nz^n\\\\
&=\frac{z}{(1-z)^2}
\end{align}$$
for $|z|<1$.
