show $ {-a\choose k} = \frac{(-1)^k (a)_{k}}{k!} $

Question

While doing an exercise and playing around in wolfram alpha I stumbled upon this identity, but I dont see how one could show it. Any hint? $(a)_{k}$ is the Pochhammer symbol so rising factorial.

$$ {-a\choose k} = \frac{(-1)^k (a)_{k}}{k!} $$

Answer 1

If we expand the LHS we get $$\frac{(-a)(-a-1)\cdots(-a-(k-1))}{k!}$$ Take out all the minus signs from the numerator – there are $k$: $$=(-1)^k\frac{a(a+1)\cdots(a+(k-1))}{k!}=\frac{(-1)^k(a)_k}{k!}$$

Answer 2

$${{-a}\choose{k}}=\frac{(-a)(-a-1)(-a-2)...(-a-k+1)}{k!}=\frac{\left[-\left(a\right)\right]\left[-\left(a+1\right)\right]\left[-\left(a+2\right)\right]...\left[-\left(a+k-1\right)\right]}{k!}$$ $$=\frac{(-1)^k(a)_k}{k!}=\frac{(-1)^k(a+k-1)!}{k!(a-1)!}=(-1)^k{{a+k-1}\choose{k}}$$