For $r \geq 1$, $k \geq 0$ both integers, I wish to show that
$$\binom{-r}{k}^{*}(-1)^{k} = \binom{r+k-1}{k}$$
(the negative binomial coefficient is the left one). By definition,
$$\binom{-r}{k}^{*}(-1)^{k} = (-1)^{k}\dfrac{(-r)(-r-1)\cdots(-r-k+1)}{k!} = (-1)^{k+1}\dfrac{r(r+1)\cdots(r+k-1)}{k!}$$
The right side clearly is $$(-1)^{k+1}\binom{r+k-1}{k}$$
but why does $(-1)^{k+1}$ "disappear"?
See, for example, under What is negative about the negative binomial distribution? here.
Best Answer
There are $k$ terms which need to be multiplied by $(-1)$ to get the desired quantity. So actually, factoring out the negatives would lead to $(-1)^{2k} = 1$ for all $k$ instead of $(-1)^{k+1}$.