How do we prove that $\int_{0}^{1}\int_{0}^{1}{\left(\ln{x}\ln{y}\right)^s\over 1-xy}dxdy=\Gamma^2(1+s)\zeta(2+2s)$? How do we prove that

$$\int_{0}^{1}\int_{0}^{1}{\left(\ln{x}\ln{y}\right)^s\over 1-xy}dxdy=\Gamma^2(1+s)\zeta(2+2s)$$

Integrate with respect to x first, let $s=1$
$$\int_{0}^{1}{\ln{y}\ln{x}\over 1-yx}dx$$
$u=\ln{x}\rightarrow xdu=dx$
$$\ln{y}\int_{0}^{\infty}{u\over y-e^{-u}}du$$
I don't think I am in the right track here, any hints please.

Hint. One may write, for  $s>-1$, $0<xy<1$,
$$
{\left(\ln{x}\ln{y}\right)^s\over 1-xy}=\sum_{n=0}^\infty (xy)^n\left(\ln{x}\ln{y}\right)^s
$$ giving
$$
\begin{align}
\int_{0}^{1}\int_{0}^{1}{\left(\ln{x}\ln{y}\right)^s\over 1-xy}dxdy&=\int_{0}^{1}\int_{0}^{1}\sum_{n=0}^\infty (xy)^n\left(\ln{x}\ln{y}\right)^sdxdy
\\\\&=\sum_{n=0}^\infty \left(\int_{0}^{1}x^n\left(-\ln{x}\right)^sdx\right)\left(\int_{0}^{1}y^n\left(-\ln{y}\right)^sdy\right)
\\\\&=\sum_{n=0}^\infty \left(\int_{0}^{1}x^n\left(-\ln{x}\right)^sdx\right)^2
\\\\&=\sum_{n=0}^\infty\left(\frac{\Gamma(s+1)}{(n+1)^{s+1}}\right)^2
\\\\&=\Gamma^2(s+1)\:\sum_{n=1}^\infty\frac{1}{n^{2s+2}}
\\\\&=\Gamma^2(1+s)\:\zeta(2+2s)
\end{align}
$$ as announced.
The interchange between $\int$ and $\sum$ may be justified by the dominated convergence theorem.
