TITLE: Prove that $\prod_{k=1}^{\infty} \big\{(1+\frac1{k})^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$
QUESTION [12 upvotes]: This result,
$$\prod_{k=1}^{\infty} \big\{\big(1+\frac1{k}\big)^{k+\frac1{2}}\big/e\big\} = \dfrac{e}{\sqrt{2\pi}}$$
is in a paper by Hirschhorn
in the current issue
of the Fibonacci Quarterly (vol. 51, no. 2).
I thought it was quite interesting in that
it shows how close
$(1+\frac1{k})^{k+\frac1{2}}$ is to $e$,
and that it would be an interesting challenge problem.
The proof there is not too difficult.
This result is a lemma in the proof of the following,
which is the principal result in the paper:
$$\prod_{k=0}^n \binom{n}{k} \sim
C^{-1}\frac{e^{n(n+2)/2}}{n^{(3n+2)/6}(2\pi)^{(2n+1)/4}}
\exp\big\{-\sum_{p\ge 1}\frac{B_{p+1}+B_{p+2}}{p(p+1)}\frac1{n^p}\big\}\text{ as }n \to \infty
$$
where
$$\begin{align}
C
&= \lim_{n \to \infty} 
\frac1{n^{1/12}}
\prod_{k=1}^n \big\{k!\big/\sqrt{2\pi k}\big(\frac{k}{e}\big)^k\big\}\\
&\approx 1.04633506677...\\
\end{align}
$$
and the $\{B_p\}$
are the Bernoulli numbers,
defined by
$$\sum_{p \ge 0} B_p\frac{x^p}{p!} = \frac{x}{e^x-1}
.$$
I don't expect anyone here to prove this,
since Hirschhorn takes over seven pages
of involved math to prove it.
Also, it was an interesting exercise in $\LaTeX$
to enter these formulae
so that they displayed exactly
(or, at least, pretty closely)
as in Hirschhorn's article.
Among other things,
I learned (after a little searching)
that a tilde (~) 
is entered as "\tilde{}".

REPLY [5 votes]: Let me put the things into what I believe to be the right context for this type of infinite products.

The proposed product appears as a part of the-so called Barnes $G$-function which is mainly characterized by the recursion relation $G(z+1)=\Gamma(z) G(z)$ and normalization $G(1)=1$. For positive integer $z$ this defines a kind of superfactorial: $$G(n)=\prod_{k=0}^{n-2}k!\tag{1}$$ 
This function has the following infinite product representation:
$$G(1+z)=\left(2\pi\right)^{\frac{z}{2}}\exp\left(-\frac{z+z^2\left(1+\gamma\right)}{2}\right)\prod_{k=1}^{\infty}
 \left(1+\frac{z}{k}\right)^k \exp\left(\frac{z^2}{2k}-z\right),\tag{2}$$
where $\gamma=-\psi(1)$ is the Euler-Mascheroni constant.
Setting $z=1$ in (2) and using (1), we obtain 
$$1=\sqrt{2\pi}e^{-1-\gamma/2}\prod_{k=1}^{\infty}
 \left(1+\frac{1}{k}\right)^k \exp\left(\frac{1}{2k}-1\right).\tag{3}$$

The formula (3) means that the required identity is equivalent to 
$$\sum_{k=1}^{\infty}\left[\frac{1}{2k}-\frac{1}{2}\ln\left(1+\frac{1}{k}\right)\right]=\frac{\gamma}{2}.$$
But this is almost the definition of $\gamma$ (note that the sum of logarithms is telescoping).