$$\prod^\infty_{k=1}k^{\frac{1}{k^2}} = \left(\frac{A ^{12}}{2\pi e^{\gamma}} \right)^{\zeta(2)}$$
Clearly by taking the log this related to the derivative of the zeta function
$$\zeta'(2) = - \log \prod^\infty_{k=1}k^{\frac{1}{k^2}} = -\sum_{k=1}^\infty \frac{\log(k)}{k^2}$$ I proved the formula using
$$\lim_{n \to \infty}\frac{G(n+1)}{(2\pi)^{n/2}n^{n^2/2-1/12} e^{-3n^2/4}} = \frac{e^{1/12}}{A}$$
The approach is too long to be written here. I think my approach is not the best because it assumes lots of things like swapping limits and integrals.
I am interested in seeing other approaches? or what was the original proof to the equation? Is there any simpler solution ?
Note that $A$ is the Glaisher-Kinkelin constant.
Ok I found another approach,
Start by
$$\zeta(s) = \lim_{m \to \infty} \left( \sum_{k=1}^{m} k^{-s} - \frac{m^{1-s}}{1-s} - \frac{m^{-s}}{2} + \frac{sm^{-s-1}}{12} \right) \ , \ \text{Re}(s) >-3.$$
Differentiate with respect to $s$
$$\zeta'(s) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k^{-s}\log(k) - \frac{m^{1-s}}{(1-s)^2} +\frac{m^{1-s}}{1-s}\log(m)+\frac{m^{-s}}{2}\log(m)\\ + \frac{m^{-s-1}}{12}-\frac{m^{-s-1}}{12}\log(m)\right) $$
Now let $s \to -1$
$$\zeta'(-1) = \lim_{m \to \infty} \left(- \sum_{k=1}^{m} k\log(k) - \frac{m^{2}}{4} +\frac{m^{2}}{2}\log(m)+\frac{m}{2}\log(m)\\ + \frac{1}{12}-\frac{1}{12}\log(m)\right) $$
Take the exponential of both sides
$$e^{\zeta'(-1)}= e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = e^{1/12}\lim_{m \to \infty} \frac{m^{m^2/2+m/2-1/12} e^{-m^2/4}}{ e^{\sum_{k=1}^{m} k\log(k)}} = \frac{ e^{1/12}}{A} $$
We conclude that
$$\zeta'(-1) = \frac{1}{12}-\log A$$
Now use the functional equation
$$\zeta(s)=2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)$$
By taking the derivative
$$\zeta'(s)=2^s\pi^{s-1}\log(2\pi)\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)+\frac{\pi}{2}2^s\pi^{s-1}\cos\left( \frac{s\pi}2\right) \Gamma(1-s)\zeta(1-s)-2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\psi(1-s)\zeta(1-s)-2^s\pi^{s-1}\sin \left( \frac{s\pi}2\right) \Gamma(1-s)\zeta'(1-s)$$
Take $s \to -1$
$$\zeta'(-1)=-2^{-1}\pi^{-2}\log(2\pi)\zeta(2)+2^{-1}\pi^{-2}\Gamma(2)\psi(2)\zeta(2)+2^{-1}\pi^{-2} \Gamma(2)\zeta'(2)$$
$$2\pi^2\zeta'(-1)=-\log(2\pi)\zeta(2)+\psi(2)\zeta(2)+ \zeta'(2)$$
Hence we have
$$\zeta'(2) = \zeta(2)(\log(2\pi)+\gamma-12\log A)$$
which implies that
$$\prod^\infty_{k=1}k^{\frac{1}{k^2}} = \left(\frac{A ^{12}}{2\pi e^{\gamma}} \right)^{\zeta(2)}$$