$\prod_n\frac{n(n+a+b)}{(n+a)(n+b)}=\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}$

For \(N\ge1\), set \[ P_N=\prod_{n=1}^N\frac{n(n+a+b)}{(n+a)(n+b)}. \] Using \(\Gamma(z+1)=z\Gamma(z)\), \[ \prod_{n=1}^N(n+a)=\frac{\Gamma(N+a+1)}{\Gamma(a+1)}, \] and similarly for the other factors. Hence \[ P_N= \frac{\Gamma(N+1)\Gamma(N+a+b+1)} {\Gamma(N+a+1)\Gamma(N+b+1)} \cdot \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}. \] The standard gamma-ratio asymptotic \[ \frac{\Gamma(N+c)}{\Gamma(N+d)}\sim N^{c-d} \] shows that the first factor tends to \(1\). Therefore \[ \boxed{ \prod_{n=1}^\infty\frac{n(n+a+b)}{(n+a)(n+b)} =\frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+1)}}. \]

For the reflection formula, Euler’s product gives \[ \frac1{\Gamma(s)}=s e^{\gamma s} \prod_{n=1}^\infty\left(1+\frac{s}{n}\right)e^{-s/n}, \] and replacing \(s\) by \(-s\) gives \[ \frac1{\Gamma(-s)}=-s e^{-\gamma s} \prod_{n=1}^\infty\left(1-\frac{s}{n}\right)e^{s/n}. \] Multiplying, \[ \frac1{\Gamma(s)\Gamma(-s)} =-s^2\prod_{n=1}^\infty\left(1-\frac{s^2}{n^2}\right). \] The sine product \[ \frac{\sin\pi s}{\pi s} =\prod_{n=1}^\infty\left(1-\frac{s^2}{n^2}\right) \] therefore yields \[ \frac1{\Gamma(s)\Gamma(-s)}=-\frac{s\sin\pi s}{\pi}. \] Since \(\Gamma(1-s)=(-s)\Gamma(-s)\), \[ \boxed{\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin\pi s}}, \] first away from the integers and then as an identity of meromorphic functions.