$\frac{1}{4-z}=\frac{1}{4}\frac{1}{1-\frac{z}{4}}$
$\frac{1}{4-z}=\frac{1}{4}\sum_{n=0}^{\infty}(\frac{z}{4})^n$
Take derivatives on both sides, then
$\frac{1}{(4-z)^2}=\sum_{n=1}^{\infty}\frac{1}{4^{n+1}}nz^{n-1}$ (Here, use the hint)
$\frac {z^2}{(4-z)^2}=\sum_{n=1}^{\infty}\frac{1}{4^{n+1}}nz^{n+1}$
$=\sum_{n=1}^{\infty}n(\frac{z}{4})^{n+1}$
OK Fiexd it.
