$$\frac{\pi^4}{120}=\sum_{(a,b) \in \mathbb{N^2_*}}'(ab)^2\prod_{n=1}^{\hat\infty}\frac{1}{n^2}=\frac{1}{2}(\sum_{n=1}^\infty\frac{1}{n^2})^2-\frac{1}{2}\sum_{n=1}^\infty\frac{1}{n^4}$$
$$\sum_{(a,b,c,...)\in \mathbb{N^k_*}}\frac{1}{(abc...)^2}=\frac{\pi^{2k}}{(2k+1)!}$$