This product is quite a famous one:
We have that the Gamma function defined as:
$\Gamma(z) = \int\limits_0^\infty e^{-t} t^{z}\frac{dt}{t}$
is convergent for $\Re(z)>0$
It is not too difficult to prove that, fo $n\in \mathbb{N}$,
$\lim\limits_{n \to \infty} \frac{\Gamma({z+n})}{ n^{z-1}n!}=1$
This with the functional equation of $\Gamma$ gives the identity:
$\Gamma(z) = \lim\limits_{n \to \infty} \frac{n^{z} n!}{\prod\limits_{v=0}^{n} (z+v)}$
Let's write
$\tag{1} \frac{1}{\Gamma_n(z)} = \frac{\prod\limits_{v=0}^{n} (z+v)}{ n^{z} n!}$
We know that
$\lim\limits_{n \to \infty} \frac{1}{\Gamma_n(z)} =\frac{1}{\Gamma(z)} $
We can write $(1)$ differently, as
$ \frac{1}{\Gamma_n(z)} = \frac{z}{n^z}{\prod\limits_{v=1}^{n} \left(\frac{z}{v}+1 \right)}$
Now, ${n^z} = {e^{z\log n}} = {e^{ - z\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n} \right)}}{e^{z\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}} \right)}}$
so we can write
$\frac{1}{{{\Gamma _n}(z)}} = z{e^{z\left( {1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log n} \right)}}\prod\limits_{v = 1}^n {\left( {\frac{z}{v} + 1} \right)} {e^{ - \frac{z}{v}}}$
Letting $n\to \infty$ gives:
$\frac{1}{{\Gamma (z)}} = z{e^{z\gamma }}\prod\limits_{v = 1}^\infty {\left( {\frac{z}{v} + 1} \right)} {e^{ - \frac{z}{v}}}$
where $\gamma\approx 0.577\dots$ is Euler's constant.
This means that
$\frac{1}{{\Gamma ( - z)}} = - z{e^{ - z\gamma }}\prod\limits_{v = 1}^\infty {\left( {1 - \frac{z}{v}} \right)} {e^{\frac{z}{v}}}$
or that
$\eqalign{ & \frac{1}{{\Gamma ( - z)}}\frac{1}{{\Gamma (z)}} = - z{e^{ - z\gamma }}z{e^{z\gamma }}\prod\limits_{v = 1}^\infty {\left( {1 - \frac{z}{v}} \right)} \left( {1 + \frac{z}{v}} \right){e^{ - \frac{z}{v}}}{e^{\frac{z}{v}}} \cr & \frac{1}{{\Gamma ( - z)}}\frac{1}{{\Gamma (z)}} = - {z^2}\prod\limits_{v = 1}^\infty {\left( {1 - \frac{z}{v}} \right)} \left( {1 + \frac{z}{v}} \right) \cr} $
But $\frac{1}{{ - z\Gamma ( - z)}} = \frac{1}{{\Gamma (1 - z)}}$ so
$\frac{1}{{\Gamma (1 - z)}}\frac{1}{{\Gamma (z)}} = z\prod\limits_{v = 1}^\infty {\left( {1 - \frac{{{z^2}}}{{{v^2}}}} \right)} $
Now put $z=\dfrac{s}{\pi}$, and your expression will appear. Note that the exponentials cancel out, and the infinite product converges everywhere.