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 $\zeta(2) = \pi^2/6$.

I would like to present you a method based on Daners' derivation of $\zeta(2)$ (which I summarised on MathStackexchange in here couple years ago).
We start defining (for all $n \in \mathbb{N}_0$)
\begin{align}
A_n & =\int_0^{\pi/2}\cos^{2n}x\;\mathrm{d}x,\\
B_n& =\int_0^{\pi/2}x^2\cos^{2n}x\;\mathrm{d}x,\\
C_n& =\int_0^{\pi/2}x^4\cos^{2n}x\;\mathrm{d}x
\end{align}
and let $\beta_n = B_n/A_n$ and $\gamma_n = C_n/A_n$.
The first integral for $A_n$ is well known and follows a reccurence relation
$$A_{n}=\frac{2n-1}{2n}A_{n-1}\tag{1}.$$
By applying per partes on the $A_n$ integral twice, we get:
$$A_n=\int_0^{\pi/2}\cos^{2n}x\;\mathrm{d}x=x\cos^{2n}x\bigg{|}_0^{\pi/2}-\frac{x^2}{2}(\cos^{2n}x)'\bigg{|}_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}x^2(\cos^{2n}x)''\;\mathrm{d}x$$
The first two terms vanish, so only the integral remains and since $(\cos^{2n}x)''=2n(2n-1)\cos^{2n-2}x-4n^2\cos^{2n}x$, we get for $n\geq 1$:
$$A_n=(2n-1)nB_{n-1}-2n^2B_{n}\tag{2}$$
Similarly, applying per partes twice on $B_n$ instead, we get
$$B_n=\frac16 (2n-1)nC_{n-1}-\frac13 n^2C_{n}\tag{3}$$
Inserting $(1)$ into $(2)$ and $(3)$ and rearranging, we get a following recurence reltions

$$\frac{1\cdot 2}{4}\frac{1}{n^2}=\beta_{n-1}-\beta_n\tag{4}$$
$$\frac{3\cdot 4}{4}\frac{\beta_n}{n^2}=\gamma_{n-1}-\gamma_n\tag{5}$$

Summing these up, we get, by the telescoping property
$$\sum_{l=1}^k\frac{1}{2l^2}=\beta_0-\beta_k \qquad \text{and} \qquad \sum_{k=1}^n\frac{3\beta_k}{k^2}=\gamma_0-\gamma_n. \tag{6}$$
Inserting one sum into another (expressing $\beta_k$), we get
$$3\beta_0\sum_{k=1}^n\frac{1}{k^2} - \frac32\sum_{k=1}^n\frac{1}{k^2}\sum_{l=1}^k\frac{1}{l^2}=\gamma_0 - \gamma_n. \tag{7}$$
Note that in general
$$2\sum_{k=1}^n a_k\sum_{l=1}^k a_l = 2\sum_{1\leq l 
\leq k \leq n}a_k a_l = \left(\sum_{k=1}^n a_k\right)^2+\sum_{k=1}^n a_k^2,$$
substituing for $a_k = 1/k^2$ into $(7)$ and rewriting $\sum_{k=1}^n 1/k^2$ as $2\beta_0 - 2\beta_n$ everywhere, we get
$$\bbox[10px,#ffd]{\sum_{k=1}^n \frac{1}{k^4} = 4\beta_0^2 - 4\beta_n^2 - \frac{4}{3}\gamma_0 + \frac{4}{3}\gamma_n.}\tag{8}$$
However, using the inequalities $x^4<(\pi/2)^2 x^2$ and $\frac{2x}{\pi}<\sin x$ valid on $(0,\frac{\pi}{2})$ and by $(1)$, we get
$$0<\gamma_{n-1} <\left(\frac{\pi}{2}\right)^2\beta_{n-1} < \frac{\int_0^{\pi/2}\sin^2x\cos^{2n-2}x}{\int_0^{\pi/2}\cos^{2n-2}x}=\frac{A_{n-1} - A_n}{A_n} = \frac{1}{2n}.$$
Applying the squeeze theorem, we get
$$\lim_{n\rightarrow \infty} \beta_n = \lim_{n\rightarrow \infty} \gamma_n = 0$$
and hence, taking the limit of $(8)$,
$$\zeta_4 = \lim_{n\rightarrow \infty} \sum_{k=1}^n\frac{1}{k^4} = 4\beta_0^2-\frac{4}{3}\gamma_0 = 4\left(\frac13 \frac{\pi^2}{2^2}\right)^2-\frac{4}{3} \frac{1}{5}\frac{\pi^4}{2^4} = \frac{\pi^4}{90}$$
This finishes the proof.
