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.

Without differentiation 
$$\sum_{n=1}^\infty nz^{n-1} = 1+2z+3z^2+\cdots \tag 1$$
$$\sum_{n=1}^\infty nz^{n} = z+2z^2+3z^3+\cdots \tag 2$$
Subtract (2) from (1)
$$\sum_{n=1}^\infty nz^{n-1}(1-z) = 1+z+z^2+\cdots$$
$$\sum_{n=1}^\infty nz^{n}= \frac{z}{(1-z)^2}$$
