Infinite product $\Gamma(\tfrac14)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}$ I saw the following infinite product on Wikipedia:
$$\Gamma\left(\tfrac14\right)=\mathrm{A}^3e^{-\mathrm{G}/\pi}2^{1/6}\sqrt{\pi}\prod_{k\ge1}\left(1-\frac1{2k}\right)^{(-1)^k k}\tag{1}$$
where $\mathrm A$ is the Glaisher constant, and $\mathrm G$ is Catalan's constant. I am looking for a proof of this product.
I haven't gotten very far with this product, other than noting that if
$$\zeta_*(s)=\sum_{k\ge1}\frac1{(1-\frac1{2k})^{(-1)^k ks}}$$
then 
$$\zeta_*'(0)=3\ln\mathrm A+\frac16\ln2+\frac12\ln\pi-\frac{\mathrm{G}}{\pi}-\ln\Gamma\left(\tfrac14\right),\tag2$$
of course, assuming that $(1)$ is true. Perhaps $(2)$ is easier to prove. Could I have some help? Thanks.
 A: A short proof.
$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$
$\displaystyle \pi^{1/2} = Q_0\left(-\frac{1}{2}\right)~~ , ~~ A^3  = 2^{7/12}Q_1\left(-\frac{1}{2}\right)^2 ~~ , ~~ e^{G/\pi} = 2^{3/4}\left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)  }\right)^2$ 
It follows:
$\displaystyle \prod\limits_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{(-1)^k k} =  \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{4k}\right)^{2k}}{\left(1-\frac{1}{4k-2}\right)^{2k-1}} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{2k}\right)^{2k-1} \left(1-\frac{1}{4k}\right)^{2k}}{ \left(1-\frac{3}{4k}\right)^{2k-1} } = $
$\displaystyle = \frac{ \prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right) n^{-1/2} }{ n^{-3/4} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right) } \left(\frac{ e^{-3n/4} n^{-9/32} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right)^k \prod\limits_{k=1}^n \left(1-\frac{1}{4k}\right)^k }{\prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right)^k  e^{-n/2} n^{-1/8} e^{-n/4} n^{-1/32} }\right)^2$
$\displaystyle \to  ~\frac{ Q_0\left(-\frac{1}{2}\right) }{ Q_0\left(-\frac{3}{4}\right) } \left(\frac{ Q_1\left(-\frac{3}{4}\right) }{ Q_1\left(-\frac{1}{2}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
$\displaystyle = 2^{-1/6} \frac{ Q_0\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{1}{2}\right) } \cdot 2^{-7/12} Q_1\left(-\frac{1}{2}\right)^{-2} \cdot 2^{3/4} \left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)}\right)^2$
$\displaystyle = \Gamma\left(\frac{1}{4}\right) 2^{-1/6} \pi^{-1/2} A^{-3} e^{G/\pi}$
A: The log of the product can be expressed in term of $\frac12\beta'(-1)-\frac12 \beta'(0)$ and  $\frac12\eta'(-1)$
where $\beta(s) = \sum_{k\ge 0} (2k+1)^{-s} (-1)^k$ 
, it is a L-function with a functional equation.
$G =\beta(2)$, $\log A = 1/12-\zeta'(-1)$
Finally $e^{\beta'(0)}$ can be expressed in term of $\Gamma(1/4)$ and $\Gamma(1/2)$ infinite products.
I don't think giving more details is useful.
