How to prove that $e^{\gamma}={e^{H_{x-1}}\over x}\prod_{n=0}^{\infty}\cdots?$ How do we show that 
$$e^{\gamma}={e^{H_{x-1}}\over x}\prod_{n=0}^{\infty}\left(\prod_{k=0}^{n}\left({k+x+1\over k+x}\right)^{(-1)^{k}{{n\choose k}}}\right)^{1\over n+2}?\tag1$$
Where $\gamma$ is the Euler-Mascheroni constant
$H_0=0$, $H_n$ is the harmonic number. 
$x\ge1$
Take the log of $(1)$
$$\gamma-H_{x-1}+\ln(x)=\sum_{n=0}^{\infty}{1\over n+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\ln\left({k+x+1\over k+x}\right)\tag2$$
Probably take $(2)$ into an integral and take it from there...?

Here's a possible start
manipulating the double sum.
$$
\begin{align}\\
S(x)&= \sum_{n=0}^{\infty}{1\over n+2}\sum_{k=0}^{n}(-1)^k{n\choose k}\ln\left({k+x+1\over k+x}\right)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x+1))-\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x)))\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n+1}(-1)^{k-1}{n\choose k-1}\ln(k+x)-\sum_{k=0}^{n}(-1)^k{n\choose k}\ln(k+x))\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n}\ln(k+x)((-1)^{k-1}{n\choose k-1}-(-1)^k{n\choose k})\\
&\quad +(-1)^n{n \choose n}\ln(n+1+x)-(-1)^0{n \choose 0}\ln(x)
)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n}\ln(k+x)(-1)^{k-1}({n\choose k-1}+{n\choose k})\\
&\quad +(-1)^n\ln(n+1+x)-\ln(x)
)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
(\sum_{k=1}^{n}\ln(k+x)(-1)^{k-1}{n+1\choose k}\\
&\quad +(-1)^n\ln(n+1+x)-\ln(x)
)\\
&=\sum_{n=0}^{\infty}{1\over n+2}
\sum_{k=0}^{n+1}\ln(k+x)(-1)^{k-1}{n+1\choose k}\\
&=-\sum_{n=0}^{\infty}{1\over n+2}
\Delta^{n+1}\ln(x)\\
\end{align}
$$
This almost looks like
a Newton series.
Not sure where to
go from here,
so I'll leave it at this
in the hope that
someone else can
take it further.
