Show that $$|(n+ib)!|= \left( \frac{\pi b}{\sinh(\pi b)}\right)^{{1}/{2}}\prod_{s=1}^{n}{(s^2+b^2)^{{1}/{2}}}$$
I have the following relations
$$\frac{\sinh(b \pi)}{b\pi} = \prod_{n=0}^{\infty}\left[1+\frac{b^2}{(n+1)^2}\right] $$
and $$ |\Gamma(1+ib)|^{-2} = \prod_{n=0}^{\infty}\left[1+\frac{b^2}{(n+1)^2}\right] $$
My attempt: $$|(n+ib)!|= |\Gamma(n+ib+1)|$$ then $$ |\Gamma(n+ib+1)|^2 = \Gamma(n+ib+1)\Gamma(n-ib+1)$$ then $$ \Gamma(n+ib+1)\Gamma(n-ib+1) = \left( \frac{\pi b}{\sinh(\pi b)} \right)\prod_{s=1}^{n}{(s^2+b^2)}$$
I do not know how to eliminate n variable in gamma function so that I can proceed using the relations listed above. Any ideas?