Showing that $\sigma=\prod_{n=1}^{\infty}(n!)^{\frac{1}{2^{n+1}}}$ Somos's quadratic recurrence constant
The Somos's Quadratic recurrence constant is defined by the sequence $g_n=ng_{n-1}$ with initial value of $ g_0= 1$
The value of $\sigma=1.661687...$
An infinite product from maths world $\sigma=\prod_{k=1}^{\infty}k^{\frac{1}{2^k}}$
We found another infinite product involving the factorial numbers by experiments on a sum calculator.
$$\sigma=\prod_{n=1}^{\infty}(n!)^{\frac{1}{2^{n+1}}}$$
Where n! is valid for non-negative integers and defined by
$n!=n(n-1)(n-2)\cdots2\cdot1$
Can somebody help us to prove this 
 A: This is an opportunity to use summation by parts:
$$
\sum _{k=0}^{n} a_kb_k = A_nb_{n}- \sum _{k=0}^{n-1} A_k(b_{k+1}-b_k) \tag1
$$  where $\displaystyle A_n:=\sum _{k=0}^{n} a_k$. Applying it with
$$
a_k=\frac1{2^k},\quad A_n=2-\frac1{2^n}, \quad b_k=\log (k!),\quad b_{k+1}-b_k=\log (k+1),
$$ gives
$$
\begin{align}
\sum _{k=0}^{n} \frac1{2^k} \log (k!) &= \left(2-\frac1{2^n}\right)\log (n!)- \sum _{k=0}^{n-1} \left(2-\frac1{2^k}\right)\log (k+1) 
\\& =\left(2-\frac1{2^n}\right)\log (n!)- \sum _{k=1}^{n} \left(2-\frac2{2^k}\right)\log k 
\\&=-\frac1{2^n}\log (n!)+2\sum _{k=1}^{n} \frac1{2^k}\log k 
\end{align}
$$ that is

$$
\sum _{k=1}^{n} \frac1{2^k}\log k=\sum _{k=0}^{n} \frac1{2^{k+1}} \log (k!)+\frac1{2^{n+1}}\log (n!) \tag2
$$ 

By letting $n \to \infty$, using $\displaystyle 0\leq\frac1{2^{n+1}}\log (n!)\leq \frac{n\log n}{2^{n+1}}$, then exponentiating both sides one gets the announced result.
A: Write $n!=\prod_{k=1}^{n} k$ and you have:
$$\begin{align}\prod_{n=1}^{\infty}(n!)^{\frac{1}{2^{n+1}}}&=\prod_{n=1}^{\infty}\prod_{k=1}^{n} k^{1/2^{n+1}}\\
&=\prod_{k=1}^{\infty}\prod_{n=k}^{\infty}k^{1/2^{n+1}}\\
&=\prod_{k=1}^{\infty}k^{\sum_{n=k}^{\infty}1/2^{n+1}}\\
&=\prod_{k=1}^{\infty}k^{1/2^k}\\
&=\sigma
\end{align}$$
