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?

Solution 1:

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...