Power series involving complex variable. Show that
$1 + \binom{m+1}{1}z + \binom{m+2}{2}z^2 +...+ \binom{m+n}{n}z^n +... = \frac{1}{(1-z)^{m+1}}$
for non-negative integers $m$ and $|z| < 1$.
 A: Hint: start with the geometric series:
$$1+z+z^2+\ldots = \frac{1}{1-z}$$
and differentiate. For example,
$$1+2 z+3 z^2+\ldots=\frac{1}{(1-z)^2}$$
$$2+(3)(2)z+(4)(3)z^2+\ldots=\frac{2}{(1-z)^3} \implies 1+\binom{3}{1} z+\binom{4}{2}z^2+\ldots=\frac{1}{(1-z)^3}$$
Differentiate $m$ times to get your result.
A: Here is a generating series/combinatorial approach.
So you know that 
$$
\frac{1}{1-z}=\sum_{n\geq 0}z^n\qquad\forall |z|<1.
$$
Recall that if two series $\sum_{n\geq 0}a_n$ and $\sum_{n\geq 0}b_n$ converge absolutely, then the Cauchy product satifies
$$
\sum_{n\geq 0}a_n\cdot \sum_{n\geq 0}b_n=\sum_{n\geq 0}c_n
$$
where the rhs converges absolutely with coefficients given by
$$
c_n=\sum_{k+l=n}a_kb_l.
$$
Applying this $m$ times to your initial power series, you get
$$
\left(\frac{1}{1-z}\right)^m=\sum_{n\geq 0} c_n(m)z^n\qquad\forall |z|<1
$$
with
$$
c_n(m)=\sum_{k_1+\ldots+k_m=n}1=\binom{m+n-1}{n}.
$$
The last equality is a well-know combinatorial formula: the number of $m$-uples $(k_1,\ldots,k_m)$ of nonnegative integers with sum equal to $n$ is the number of ways of picking $m-1$ balls among $m+n-1$ balls. That is $\binom{m+n-1}{n}$.
