There was not problem with the link whatsoever...
My first approach:
Use two expansions of the function
\frac{\sin x}{x}
around zero,one being the Taylor series:
\frac{\sin x}{x}=1-\frac{x^{2}}{3!}+\frac{x^{4}}{5!}-\frac{x^{6}}{7!} +... (1)
,and the other the famous Euler's sine product:
\frac{\sin x}{x}=(1-\frac{x^{2}}{\pi^{2}})(1-\frac{x^{2}}{4\pi^{2}})(1-\frac{x^{2}}{9\pi^{2}})... (2)
Expand (2) and write it:
\frac{\sin x}{x}=1-\frac{x^{2}}{\pi^{2}}\sum_{n=1}^{\infty}\frac{1}{n^{2}}+\frac{x^{4}}{\pi^{4}}\sum_{n,k=1;n\neq k}^{\infty} \frac{1}{n^{2}k^{2}}-\frac{x^{6}}{\pi^{6}}\sum_{n,k,l=1;n\neq k\neq l}^{\infty} \frac{1}{n^{2}k^{2}l^{2}}+... (3)
Equate (1) and (3) to find:
\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6} (4)
\sum_{n,k=1;n\neq k}^{\infty} \frac{1}{n^{2}k^{2}}=\frac{\pi^{4}}{120} (5)
Square relation (4) to find:
(\sum_{n=1}^{\infty}\frac{1}{n^{2}})^{2}=\sum_{n=1}^{\infty}\frac{1}{n^{4}}+2\sum_{n,k=1;n\neq k}^{\infty} \frac{1}{n^{2}k^{2}}=\frac{\pi^{4}}{36} (6)
From (5) & (6) u find simply that:
\sum_{n=1}^{\infty} \frac{1}{n^{4}}=\frac{\pi^{4}}{36}-2\frac{\pi^{4}}{120}=\frac{\pi^{4}}{90}
I don't personly like this method,coz i find working with sums rather difficult and unintuitive.My second method is more eleborate (requires some simple integrations),but i find more intuitive.
Daniel.
It reminds me of the "Terminator" series in which the series for robots kept popping up from time to time... 
