Prove complex equation Prove the following $$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$$ 
for any integer n greater than 0. DO NOT USE ....
I believe that I can use mathematical induction. Any help would be greatly appreciated.

You want to show that
$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$
Multiply by 
$z-1$ and it becomes
$\frac{1}{z^n}
= 1 - (z-1)\sum_{k=1}^n\frac{1}{z^k}
$
or
$1-\frac{1}{z^n}
=  (z-1)\sum_{k=1}^n\frac{1}{z^k}
$.
But
$\begin{array}\\
(z-1)\sum_{k=1}^n\frac{1}{z^k}
&=z\sum_{k=1}^n\frac{1}{z^k}-\sum_{k=1}^n\frac{1}{z^k}\\
&=\sum_{k=0}^{n-1}\frac{1}{z^k}-\sum_{k=1}^n\frac{1}{z^k}\\
&=1-\frac1{z^n}\\
\end{array}
$
QED
