Note that: $$\frac{z^2}{(z+i)(z-i)^2}\equiv \frac{z^2(z+i)}{(z^2+1)^2}$$
This means, that I would need to only find the power series of $\displaystyle\frac1{(z^2+1)^2}$.
We have: $$\frac1{1-x}\equiv\sum_{n\mathop=0}^\infty x^n$$
Taking derivative of both sides: $$\frac{-1}{(1-x)^2}\equiv\sum_{n\mathop=0}^\infty nx^{n-1}\equiv\sum_{n\mathop=0}^\infty (n+1)x^n$$
Substituting $x=-z^2$: $$\frac{-1}{(1+z^2)^2}\equiv\sum_{n\mathop=0}^\infty (n+1)(-z^2)^n\equiv\sum_{n\mathop=0}^\infty (n+1)(iz)^{2n}$$
Multiplying both sides by $-z^2(z+i)$: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv(z+i)\sum_{n\mathop=0}^\infty (n+1)(iz)^{2n+2}\equiv(z+i)\sum_{n\mathop=1}^\infty n(iz)^{2n}$$
Distributing: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv\sum_{n\mathop=1}^\infty i^{2n}nz^{2n+1}+\sum_{n\mathop=1}^\infty i^{2n+1}nz^{2n}$$
Combining: $$\frac{z^2(z+i)}{(1+z^2)^2}\equiv\sum_{n\mathop=1}^\infty (-i)^{2n}nz^{2n+1}+\sum_{n\mathop=1}^\infty (-i)^{2n-1}nz^{2n}$$
$\displaystyle\equiv\sum_{n\mathop=\mbox{odd}}^\infty (-i)^{n+1}\frac{n+1}2z^{n+2}+\sum_{n\mathop=\mbox{even from 0}}^\infty (-i)^{n+1}\frac {n+2}2z^{n+2}$
$\displaystyle\equiv\sum_{n\mathop=\mbox{odd}}^\infty (-i)^{n+1}\left\lceil\frac n2+1\right\rceil z^{n+2}+\sum_{n\mathop=\mbox{even from 0}}^\infty (-i)^{n+1}\frac {n+2}2z^{n+2}$
$\displaystyle\equiv\sum_{n\mathop=0}^\infty (-i)^{n+1}\left\lceil\frac n2+1\right\rceil z^{n+2}$
$\displaystyle\equiv\sum_{n\mathop=2}^\infty (-i)^{n-1}\left\lceil\frac n2\right\rceil z^n$
Note that\begin{align}\frac1{2-z}&=\frac1{-1-(z-3)}\\&=-\frac1{1+(z-3)}\\&=-\sum_{n=0}^\infty(-1)^n(z-3)^n\text{ (if $\lvert z-3\rvert<1$)}\\&=\sum_{n=0}^\infty(-1)^{n+1}(z-3)^n\end{align}and that therefore\begin{align}\frac1{(2-z)^2}&=\left(\frac1{2-z}\right)'\\&=\left(\sum_{n=0}^\infty(-1)^{n+1}(z-3)^n\right)'\\&=\sum_{n=1}^\infty(-1)^{n+1}n(z-3)^{n-1}\\&=\sum_{n=0}^\infty(-1)^n(n+1)(z-3)^n.\end{align}
Best Answer
Based on the power series for $e^{z}$ around $0$, one gets the desired result as follows:
\begin{align*} e^{z} = 1 + z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \ldots & \Rightarrow e^{z} - 1 = z + \frac{z^{2}}{2!} + \frac{z^{3}}{3!} + \ldots \Rightarrow \frac{e^{z} - 1}{z} = 1 + \frac{z}{2!} + \frac{z^{2}}{3!} + \ldots \end{align*}
Hopefully this helps!