Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$.  For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\zeta(4) = \sum_{n=1}^{\infty} \frac{1}{n^4} = \frac{\pi^4}{90}?$$
I already know some proofs that give all values of $\zeta(n)$ for positive even integers $n$ (like #7 on Robin Chapman's list or Qiaochu Yuan's answer in the linked question).  I'm not so much interested in those kinds of proofs as I am those that are specifically for $\zeta(4)$.
I would be particularly interested in a proof that isn't an adaption of one that $\zeta(2) = \pi^2/6$.

By induction we can easily prove that for any nonnegative real numbers $a_k$
$$1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j-\sum_{1\le i<j<k\le n}a_ia_ja_k\le\prod_{k=1}^n(1-a_k)\le1-\sum_{k=1}^na_k+\sum_{1\le i<j\le n}a_ia_j$$
Taking $a_k=\frac{x^2}{k^2\pi^2}$,we get
$$1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2-\frac{x^6}{\pi^6}\frac{\zeta_n(2)^3-\zeta_n(6)}{6}\le\prod_{k=1}^n(1-\frac{x^2}{k^2\pi^2})\le1-\frac{x^2}{\pi^2}\zeta_n(2)+\frac{x^4}{\pi^4}\frac{\zeta_n(2)^2-\zeta_n(4)}2$$
Since $\prod\limits_{k=1}^{\infty}(1-\frac{x^2}{k^2\pi^2})=\frac{\sin(x)}x$, by taking $n\to\infty$ 
$$1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le1-\frac{x^2}{\pi^2}\zeta(2)+\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$
subtraciting $1-\frac{x^2}{\pi^2}\zeta_n(2)$
$$\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2-\frac{x^6}{\pi^6}\frac{\zeta(2)^3-\zeta(6)}{6}\le\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\le\frac{x^4}{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2$$
dividing by $x^4$ and putting $x=0$ we get
$$\frac1{\pi^4}\frac{\zeta(2)^2-\zeta(4)}2=\frac1{5!}$$
and this follows that $\zeta(4)=\frac{\pi^4}{90}$.
