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.

Another way:
$$S(z)=z+2z^2+3z^3+4z^4+\cdots\Rightarrow $$ $$zS(z)=z^2+2z^3+3z^4+4z^5+\cdots\Rightarrow$$ $$\Rightarrow S(z)-zS(z)=S(z)(1-z)=z+z^2+z^3+\cdots=-1+(1+z+z^2+z^3+\cdots)$$ $$=-1+\frac{1}{1-z}=\frac{z}{1-z}\Rightarrow S(z)=\frac{z}{(1-z)^2}\Rightarrow \sum_{n=1}^{+\infty}nz^n=\frac{z}{(1-z)^2}\quad (|z|<1).$$ 
EDIT: I didn't see Zaid Alyafeai answer.
