binomial identity: $\sum_{k=x+y}^{\infty}\binom{k-1}{y-1}\binom{k-y}{x}u^k = \binom{x+y-1}{y-1}\left(\frac{u}{1-u}\right)^{x+y}$? I met a problem which gave me the left part, and I can compute left part and get right part by Mathematica. However, I don't know how to prove:
$$\sum_{k=x+y}^{\infty}\binom{k-1}{y-1}\binom{k-y}{x}u^k = \binom{x+y-1}{y-1}\left(\frac{u}{1-u}\right)^{x+y}$$
with $x, y \in \mathbb{Z}, x \ge 0, y \ge 1, 0 \le u < 1$.
My Questions:

*

*How to prove above binomial identity?


*Is there simple argument behind it? Since it's quite simple, maybe we can construct two equivalent counting processes.
 A: As explained in this thread, the taylor polynomial of
$$\left(\frac{1}{1-u}\right)^{x+y} = \sum_{k=0}^\infty {k+x+y-1 \choose x+y-1 } u^k$$
So
\begin{align}\binom{x+y-1}{y-1}\left(\frac{u}{1-u}\right)^{x+y} 
&= \sum_{k=0}^\infty \binom{x+y-1}{y-1}{k+x+y-1 \choose x+y-1 } u^{k+x+y} 
\\
&= \sum_{k=0}^\infty \frac{(k+x+y-1)!}{x!(y-1)!k!} u^{k+x+y}
\\
&=  \sum_{k=x+y}^\infty \frac{(k-1)!}{x!(y-1)!(k-x-y)!} u^{k} 
\\
&= \sum_{k=x+y}^\infty \binom{k-1}{y-1}{k-y \choose x }u^{k} 
\end{align}
A: Here is another answer and in the end I can use some help.
Let first term in left hand sum be $s_k$.
Then $s_k = \binom{x + y - 1}{y-1} u^{x+y}$
Then $s_{k+1} = \binom{x + y}{y-1} \binom{x+1}{x} u^{x+y+1}$
After simplification we get
$s_{k+1} = \frac{k}{k+1-x-y} s_k$
And we get
$s_{k+2} = \frac{k+1}{k+2-x-y} s_{k+1} = \frac{(k+1)k}{(k+2-x-y)(k+1-x-y)}$
Let $x+y = n$
So $S = \binom{x + y - 1}{y-1} u^{x+y} \left(1 + \binom{n}{1}u + \binom{n+1}{2}u^2 + ... \right)$
The last term is exactly equal to $\left(\frac{1}{1-u}\right)^n$
After some tinkering I needed to prove:
The $(n-1)th $ derivative of $\frac{u^{n-1}}{(n-1)!(1-u)} = \left(\frac{1}{1-u}\right)^n$
I know this is true but no idea how to prove it.
Thanks
