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...?
 A: This can indeed be solved by Euler series acceleration. Notice first that
$$ f(s) := \log\left(1 + \frac{1}{x+s}\right) = \int_{0}^{\infty} \frac{1 - e^{-u}}{u} \, e^{-su} \, du. $$
Then its $n$-fold forward difference satisfies
$$ (-1)^n \Delta^n f (s)
= \sum_{k=0}^{n} (-1)^{k}\binom{n}{k} f(s+k)
= \int_{0}^{\infty} \frac{(1 - e^{-u})^{n+1}}{u}\, e^{-su} \, du, \tag{1} $$
which is monotone decreasing to $0$ as $s \to \infty$. Next we refer to the following version of the Euler series acceleration: (See my blog posting for a proof, for instance.)

Theorem. If $\sum_{n=0}^{\infty} a_n z^n$ has radius of convergence $\geq 1$. Then for any $t \in [0, 1)$,
$$ \sum_{n=0}^{\infty} (-1)^n a_n t^n = \frac{1}{1+t} \sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 \left( \frac{t}{1+t}\right)^n $$
holds.

In order to apply this to our case, set $z = t/(1+t)$ and plug $a_n = f(x+n)$, where $x > 0$. Then for $t \in [0, 1)$ it follows that
\begin{align*}
\sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 z^{n+1}
&= \sum_{n=0}^{\infty} (-1)^n a_n t^{n+1} \\
&= \sum_{n=0}^{\infty} (-1)^n t^{n+1} \int_{0}^{\infty} \frac{1 - e^{-u}}{u} \, e^{-(x+n)u} \, du \\
&= \int_{0}^{\infty} \frac{1 - e^{-u}}{u} \frac{t e^{-xu}}{1 + t e^{-u}} \, du \\
&= \int_{0}^{\infty} \frac{z (1 - e^{-u}) e^{-xu}}{1 - z(1 - e^{-u})} \, \frac{du}{u}.
\end{align*}
Although this manipulation holds only when $z \in [0, \frac{1}{2})$ (mainly because of the restriction of the theorem above), now both sides define a holomorphic function on $\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}$ and hence the identity extends to all of $\mathbb{D}$ by the principle of analytic continuation.
Moreover, the left-hand side converges even when $z = 1$ due to the alternating series test combined with our previous remark on $\text{(1)}$. So by the Abel's theorem,
\begin{align*}
S(x)
&:= \sum_{n=0}^{\infty} \frac{1}{n+2} \sum_{k=0}^{n} (-1)^k \binom{n}{k} \log\left(1+\frac{1}{k+x}\right) \\
&= \sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 \int_{0}^{1} z^{n+1} \, dz \\
&= \int_{0}^{1} \sum_{n=0}^{\infty} (-1)^n (\Delta^n a)_0 z^{n+1} \, dz \\
&= \int_{0}^{1} \left( \int_{0}^{\infty} \frac{z (1 - e^{-u}) e^{-xu}}{1 - z(1 - e^{-u})} \, \frac{du}{u} \right) \, dz \\
&= \int_{0}^{\infty} \left( \frac{1}{1 - e^{-u}} - \frac{1}{u} \right)  e^{-xu} \, du.
\end{align*}
Computing this integral is not hard; differentiate both sides to obtain $S'(x) = \frac{1}{x} - \psi'(x) $ and then utilize the condition $\lim_{x\to\infty}S(x) = 0$ to obtain
$$ S(x)
= \log x - \psi (x)
= \log x + \gamma - H_{x-1}. $$
A: 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.
