Induction help for a Combinatorics problem. I've been asked to prove the following problem with induction and I'm not sure how proceed.
$\textbf{Given}$ $\frac{1}{1-z}=\sum^\infty_{k=0}z^k, |z|<1$
$\textbf{Prove}$ $\forall n\in\mathbb{Z}:n\ge1$, $\frac{1}{(1-z)^n}=\sum^{\infty}_{k=0}\binom{n+k-1}{k}z^k, |z|<1$
Since the base case $n=1$ is given, I must show that $\frac{1}{(1-z)^{n+1}}=\sum^{\infty}_{h=0}\binom{n+h}{h}z^h$

\begin{align*}
\frac{1}{(1-z)^{n+1}} &=\frac{1}{(1-z)^n}\frac{1}{1-z}\\
&=\left( \sum^\infty_{i=0}\binom{n+i-1}{i}z^i \right)\left( \sum^\infty_{j=0}z^j \right)\\
&=\sum^\infty_{i=0}\sum^\infty_{j=0}\binom{n+i-1}{i}z^{i+j}\\
&= ?\\
&=\sum^{\infty}_{h=0}\binom{n+h}{h}z^h, h=i+j
\end{align*}
Any help concerning how to solve this is welcome.
 A: If $i+j=h$, there are $h+1$ values of $i$ for which this is possible, namely $0, \ldots, h$. $j$ is uniquely determined by the value of $i$ and $h$. So,
$$\sum_{i=0}^\infty\sum_{j=0}^\infty\binom{n+i−1}{i}z^i = \sum_{h=0}^\infty z^h\sum_{i=0}^h\binom{n+i−1}{i}$$
So what needs to be shown is that 
$$\sum_{i=0}^h\binom{n+i−1}{i} = \binom{n+h}{h}$$
i.e. $$\binom{n−1}{0} + \binom{n}{1} +\ldots \binom{n+h−1}{h} = \binom{n+h}{h}$$
If we write the first term as $\binom{n}{0}$ instead of $\binom{n−1}{0}$ and repeatedly use the Pascal's identity:
$$\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$$
the correctness of the sum becomes clear.
The above identity which needed to be shown can also be evaluated by a double counting argument: The RHS is the number of $n$-size collections from $n+h$ elements. The LHS counts the same thing, but divided into cases. $\binom{n+i−1}{i} = \binom{n+i−1}{n-1}$ counts the numbers of $n$-size collections where the maximum elements is $n+i$, and clearly on summing the two are counting the same quantity.
