an identity involving products of ratios of gamma

Question

I found the following identity from a special functions book while I have no idea proving it.

Suppose $\sum_{m=1}^k(a_m-b_m)=0$, $$\prod_{n=1}^\infty\frac{(n-a_1)\cdots(n-a_k)}{(n-b_1)\cdots(n-b_k)}=\prod_{m=1}^k\frac{\Gamma(1-b_m)}{\Gamma(1-a_m)},$$ where $\Gamma(z)$ is the gamma function.

Any tips regarding how to derive this? Thanks in advance.

Answer 1

Immediately follows from infinite product representations of $\Gamma$, say $$\Gamma(1-z)=\prod_{n=1}^{\infty}\frac{(1+1/n)^{-z}}{1-z/n}.$$