How do I prove the negative binomial identity? I'm having trouble proving the negative binomial identity ${r\choose k} = (-1)^k{k-r-1\choose k}$.  Here's what I've got so far:
I know that ${k-r-1\choose k} = {(k-r-1)!\over k!(-r-1)!}$, and the numerator when expanded out has k terms.  If we distribute the $(-1)^k$ across each of those terms we end up with the numerator $(r-k+1)(r-k+2)\cdots(r-1)(r)$, which is exactly the numerator that we want for $r\choose k$.  However, the denominator for $r\choose k$ is $k!(r-k)!$ rather than $k!(-r-1)!$  What am I missing here?
 A: Hint:
Writing out the terms,
$${r \choose k}={(r-k+1)(r-k+2)\cdots(r-1)r\over 1\cdot 2\cdot 3\cdots k}$$
What terms of "r choose k" are not present due to being canceled out?  Note that the count continues from $(r-1)\cdot r\cdots$ instead of from the other side...
A: $\displaystyle{\Gamma\left(z\right)\mbox{: Gamma function}.\qquad\qquad}
$Euler Reflection Formula: 
$\displaystyle{\quad%
\Gamma\left(z\right)\,\Gamma\left(1 - z\right)
=
{\pi \over \sin\left(\pi z\right)}
}$
\begin{align}
{r \choose k}
&=
{r! \over k!\left(r - k\right)!}
=
{1 \over k!}\,{\Gamma\left(r + 1\right) \over \Gamma\left(r - k + 1\right)}
\tag{1}
\end{align}
\begin{align}
{\Gamma\left(r + 1\right) \over \Gamma\left(r - k + 1\right)}
&=
\overbrace{%
{\pi
 \over
 \sin\left(\pi\left[r + 1\right]\right)
 \Gamma\left(1 - \left[1 + r\right]\right)}}
^{\displaystyle{\Gamma\left(r\ + 1\right)}}\quad
\overbrace{%
{\sin\left(\pi\left[r - k + 1\right]\right)
 \Gamma\left(1 - \left[r - k + 1\right]\right) \over \pi}}
^{\displaystyle{1 \over \Gamma\left(r\ - k\ +   1\right)}}
\\[3mm]&=
{1 \over -\sin\left(\pi r\right)\Gamma\left(-r\right)}
\left\{\vphantom{\Large A}%
-\sin\left(\pi\left[r - k\right]\right)\Gamma\left(k - r\right)\right\}
=
{\sin\left(\pi\left[r - k\right]\right) \over \sin\left(\pi r\right)}\,
{\left(k - r - 1\right)! \over \left(-r - 1\right)!}
\end{align}
By replacing this result in $\left(1\right)$, we get
\begin{equation}
{r \choose k}
=
{\sin\left(\pi\left[r - k\right]\right) \over \sin\left(\pi r\right)}\,
{\left(k - r - 1\right)! \over k!\left(-r - 1\right)!}
\quad\Longrightarrow\quad
{r \choose k}
=
{\sin\left(\pi\left[r - k\right]\right) \over \sin\left(\pi r\right)}\,
{k - r - 1 \choose k}
\tag{2}
\end{equation}
When $k$ is an integer,
$\displaystyle{%
{\sin\left(\pi\left[r - k\right]\right) \over \sin\left(\pi r\right)}\,
}
=
\left(-1\right)^{k}$. Then, $\left(2\right)$ becomes

For $r, k \in {\mathbb Z};\quad r \leq -1$:
$$\color{#ff0000}{\large%
{r \choose k}
\color{#000000}{\ =\ }
\left(-1\right)^{k}{k - r - 1 \choose k}\,,
\qquad
\color{#000000}{k \geq 0\,,\quad r \leq -1}}
\tag{3}
$$

When $k \leq r \leq -1$, we use formula $\left(3\right)$ as follows:
$$
\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{r \choose k}
=
{r \choose r - k}
=
\left(-1\right)^{r - k}
{\left[r - k\right] - r - 1 \choose r - k}
=
\left(-1\right)^{r - k}{-k - 1 \choose r - k}\,,
\qquad
k \leq r \leq -1
\tag{4}
$$

Otherwise, $\displaystyle{{r \choose k} = 0}$ since $\Gamma\left(z\right)$ has poles at $z = 0, -1, -2, \ldots$.

A detailed account of this topic is given in
http://mathworld.wolfram.com/BinomialCoefficient.html
