sequences and series - 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.

Answer

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 $|z|<1$.