Absolute Convergence of an Infinite Product based on Weierstrass's Factor Theorem I am trying to show that $\left\{ \left( 1-\dfrac {z} {\pi }\right) e^{\left( \dfrac {z} {\pi }\right) }\right\} \left\{ \left( 1+\dfrac {z} {\pi }\right) e^{\left( -\dfrac {z} {\pi }\right) }\right\}   \left\{ \left( 1-\dfrac {z} {2\pi }\right) e^{\left( \dfrac {z} {2\pi }\right) }\right\} \left\{ \left( 1+\dfrac {z} {2\pi }\right) e^{\left( -\dfrac {z} {2\pi }\right) }\right\}\ldots $ Converges absolutely.
So the general term is of the form $\left( 1\pm \dfrac {z} {m\pi }\right) e^{\pm \dfrac {z} {m\pi }}$ I think it's the $e^{\pm \dfrac {z} {m\pi }}$ which is giving me grief. Any help would be much appreciated.
Edit:
Is multiplying the $2m -1 $ and the $2m$ terms and canceling out the $e^{\pm \dfrac {z} {m\pi }}$ parts and $\left( 1^2\ - (\dfrac {z} {m\pi })^2\right)$ terms be allowed or would that run the risk of altering the value of the product ?

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.
