I wanna show that $\sum{\frac{1}{k^{4}}}=\frac{\pi^{4}}{90}$. For this, I know that $$\sin(z)=z-\dfrac{z^{3}}{3!}+\dfrac{z^{5}}{5!}-\dfrac{z^{7}}{7!}+\cdots$$
On the other hand, also know that
$$\sin(z)=z\prod_{k=1}^{\infty}{\left(1-\dfrac{z^{2}}{\pi^{2}k^{2}}\right)}$$ $$=z\left(1-z^{2}(\cdots)+\dfrac{z^{4}}{\pi^{4}}\left(\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\cdots+\dfrac{1}{2^{2}3^{2}}+\dfrac{1}{2^{2}4^{2}}+\cdots+\dfrac{1}{3^{2}4^{2}}+\cdots\right)\right)$$
Therefore, I can compare the coefficients of the term $z^{5}$, i.e.,
$$\dfrac{1}{\pi^{4}}\left(\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\cdots+\dfrac{1}{2^{2}3^{2}}+\dfrac{1}{2^{2}4^{2}}+\cdots+\dfrac{1}{3^{2}4^{2}}+\cdots\right)=\dfrac{1}{5!}$$ Then, I should prove a relation between $\left(\right)$ and $\sum{\frac{1}{k^{4}}}$
So, if define the parenthesis as $S$, then
$$S=\dfrac{1}{2^{2}}\left(1+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{6^{2}}+\cdots\right)+\dfrac{1}{3^{2}}\left(1+\dfrac{1}{4^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{6^{2}}+\dfrac{1}{7^{2}}+\cdots\right)+\dfrac{1}{4^{2}}\left(1+\dfrac{1}{5^{2}}+\dfrac{1}{6^{2}}+\dfrac{1}{7^{2}}+\dfrac{1}{8^{2}}+\cdots\right)+\dfrac{1}{5^{2}}\left(1+\dfrac{1}{6^{2}}+\dfrac{1}{7^{2}}+\cdots\right)+\cdots$$ $$=\dfrac{1}{2^{2}}\left(\dfrac{\pi^{2}}{6}-\dfrac{1}{2^{2}}\right)+\dfrac{1}{3^{2}}\left(\dfrac{\pi^{2}}{6}-\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}\right)+\dfrac{1}{4^{2}}\left(\dfrac{\pi^{2}}{6}-\dfrac{1}{2^{2}}-\dfrac{1}{3^{2}}-\dfrac{1}{4^{2}}\right)+\cdots$$ $$=\left(\dfrac{\pi^{4}}{36}-\dfrac{\pi^{2}}{6}\right)-\sum_{k=2}^{\infty}{\dfrac{1}{k^{4}}}-\left(\dfrac{1}{3^{2}2^{2}}+\dfrac{1}{4^{2}2^{2}}+\dfrac{1}{4^{2}3^{2}}+\dfrac{1}{5^{2}2^{2}}+\dfrac{1}{5^{2}3^{2}}+\dfrac{1}{5^{2}4^{2}}+\dfrac{1}{6^{2}2^{2}}+\cdots\right)$$
Because the sum $\dfrac{\pi^{2}}{6}\left(\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{6^{2}}+\cdots\right)$ is equal to $$\dfrac{\pi^{2}}{6}\left(1+\dfrac{1}{2^{2}}+\dfrac{1}{3^{2}}+\dfrac{1}{4^{2}}+\dfrac{1}{5^{2}}+\dfrac{1}{6^{2}}+\cdots-1\right)=\dfrac{\pi^{4}}{36}-\dfrac{\pi^{2}}{6}$$
Finally, How find the sum of $$\left(\dfrac{1}{3^{2}2^{2}}+\dfrac{1}{4^{2}2^{2}}+\dfrac{1}{4^{2}3^{2}}+\dfrac{1}{5^{2}2^{2}}+\dfrac{1}{5^{2}3^{2}}+\dfrac{1}{5^{2}4^{2}}+\dfrac{1}{6^{2}2^{2}}+\cdots\right)$$
