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.
 A: 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
A: And here's a proof by induction.
It is true for $n=0$
when it is
$\frac1{1-z}
=
\frac1{1-z}
$.
Suppose it is true for $n$.
We want to show that
it is true for $n+1$,
or that
$\frac{1}{z-1}*\frac{1}{z^{n+1}}
= \dfrac{1}{z-1} - \sum_{k=1}^{n+1}\frac{1}{z^k}
$.
Using the induction hypothesis that
$\frac{1}{z-1}*\frac{1}{z^{n}}
= \dfrac{1}{z-1} - \sum_{k=1}^{n}\frac{1}{z^k}
$
$\begin{array}\\
\dfrac{1}{z-1} - \sum_{k=1}^{n+1}\frac{1}{z^k}
&=\dfrac{1}{z-1} - \left(\sum_{k=1}^{n}\frac{1}{z^k}+\frac1{z^{n+1}}\right)\\
&=\left(\dfrac{1}{z-1} - \sum_{k=1}^{n}\frac{1}{z^k}\right)-\frac1{z^{n+1}}\\
&=\left(\frac{1}{z-1}*\frac{1}{z^{n}}\right)-\frac1{z^{n+1}}
\text{ (induction hypothesis used here)}
\\
&=\frac{z-(z-1)}{(z-1)z^{n+1}}\\
&=\frac{1}{(z-1)z^{n+1}}\\
\end{array}
$
and the result is true for $n+1$.
