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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.

Clarinetist
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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 $(-1)^{k+1}$.

Clarinetist
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    Yes, this is correct. You could instead observe that $$\binom{-r}k =\frac1{k!}\prod_{i=0}^{k-1}(-r-i) =\frac{(-1)^k}{k!}\prod_{i=0}^{k-1}(r+i) =\frac{(-1)^k}{k!}\cdot\frac{(r+k-1)!}{(r-1)!} =(-1)^k\binom{r+k-1}k$$ and then multiply both sides by $(-1)^k$. – Brian M. Scott Jan 20 '16 at 23:44