<s>[INST] Formula for $\prod\limits_{k=-\infty}^\infty \frac{1}{e}\left(1+\frac{1}{k+z}\right)^{k+z+\frac{1}{2}}$ Let $\displaystyle ~(-1)^z:=e^{i\pi z}~$ for $~z\in\mathbb{C} ~\land~\Re(z)\geq 0~$ .
$\displaystyle A(z):=\prod\limits_{k=0}^\infty \frac{1}{e}\left(1+\frac{1}{k+z}\right)^{k+z+\frac{1}{2}}~$ for $~z\in\mathbb{C}\setminus\mathbb{Z}~$ .
From that follows immediately $\displaystyle ~A(-z)\,A(z)=\prod\limits_{k=-\infty}^\infty \frac{1}{e}\left(1+\frac{1}{k+z}\right)^{k+z+\frac{1}{2}}~$ .
Question: $~$ How can we proof  

$$A(-z)\,A(z)=\frac{1}{1-e^{-i2\pi z}} ~~, \hspace{5mm}\Re(z)\geq 0 \hspace{1cm} ?$$


Background: 
Proof of the well-known Euler’s Infinite Product for the Sine using the Stirling formula.
$\displaystyle F(z) := \Gamma(z+1) = \left(\frac{z}{e}\right)^z\sqrt{2\pi z}\,A(z)$ 
Formal extension:  $\enspace\displaystyle F(-z):=\left(-\frac{z}{e}\right)^{-z}\sqrt{-2\pi z}\,A(-z)$
We get $\enspace\displaystyle F(-z) = e^{i\pi\left(-z+\frac{1}{2}\right)}\left(\frac{z}{e}\right)^{-z}\sqrt{2\pi z}\,A(-z) = \frac{e^{i\pi\left(-z+\frac{1}{2}\right)}}{1-e^{-i2\pi z}}\left(\frac{e}{z}\right)^z\frac{\sqrt{2\pi z}}{A(z)} = 
\frac{\sqrt{2\pi z}}{2A(z)\sin(\pi z)}\left(\frac{e}{z}\right)^z$ 
and therefore $\enspace\displaystyle F(-z)\,F(z) = \frac{\sqrt{2\pi z}}{2\,A(z)\sin(\pi z)}\left(\frac{e}{z}\right)^z \cdot \left(\frac{z}{e}\right)^z\sqrt{2\pi z}\,A(z) = \frac{\pi z}{\sin(\pi z)}~$ . 

EDIT:
After a discussion with Diger, I think it's a problem to use the definition of $A(z)$ for $A(-z)$ . It comes from the interpretation of $(-1)^{-z}$ and $(\frac{1}{-1})^z$ . It makes sense to define:
$$A(z):=\prod\limits_{k=0}^\infty \frac{1}{e}\frac{(k+z+1)^{k+z+\frac{1}{2}}}{(k+z)^{k+z+\frac{1}{2}}}$$
EDIT 2: $~~$Solution
J. R. Quine, S. H. Heydari and R. Y. Song are discussing Zeta Regularized Products here.
On page 226, example $10$, is shown how my question could be answered.  
 [/INST] You can calculate $\log A(z)$ explicitly, by diferentiating with respect to $z$ two times and carrying out the summation.
You'll then obtain $$\frac{{\rm d}^2}{{\rm d}z^2} \, \log A(z) = \Psi'(z) - \frac{1}{z} - \frac{1}{2z^2} \\ 
\Longrightarrow \quad \log A(z) = \log \Gamma(z+1) + z - z\log z - \log \sqrt{z} + c_1 z + c_2 \, .$$
By comparing the limit value for $z \rightarrow \infty$ of the original sum (which is $0$), you will need $c_1=0$ and $c_2=-\log\sqrt{2\pi}$, so $$\log A(z) = \log \Gamma(z+1) + z - z\log z - \log \sqrt{2\pi z} \\ A(z) = \frac{\Gamma(z+1)}{\sqrt{2\pi z}} \left(\frac{e}{z}\right)^z \, .$$ I presume you can use this backwards to prove your equality. Note that the exponent $1/2$ is crucial in the sum/product. For any other value the sum/product diverges.

To be honest I still don't know what you precisely want :-(. Have you tried
\begin{align}
&\qquad \prod_{k=-N}^N \frac{1}{e} \, \frac{(k+z+1)^{k+z+1/2}}{(k+z)^{k+z+1/2}} \\
&=(1+N+z)^{N+z+1/2} (z-N)^{N-z+1/2} \frac{e^{-2N-1}}{z} \prod_{k=1}^N \frac{1}{z^2-k^2} \\
&=\left(1+\frac{z+1}{N}\right)^{N+z+1/2} N^{N+z+1/2} \cdot \left(1-\frac{z}{N}\right)^{N-z+1/2} N^{N-z+1/2} \\ &\quad \cdot (-1)^{1/2-z} \, \frac{e^{-2N-1}}{z} \cdot \frac{1}{N!^2} \prod_{k=1}^N \frac{1}{1-\left(\frac{z}{k}\right)^2} \\
&\sim \frac{(-1)^{1/2-z}}{2\pi z}  \prod_{k=1}^\infty \frac{1}{1-\left(\frac{z}{k}\right)^2} \\
&= \frac{i e^{-i\pi z}}{2\sin(\pi z)} = \frac{1}{1-e^{i2\pi z}} \, ?
\end{align}
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