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

On the proofs of the Basel problem page there is one "peculiar" answer which counts the number of integer solutions to $a^2+b^2+c^2+d^2=n$ as a function of $n$. This is an analog to that answer. Let $r_8(n)$ denote the number of integer solutions to $$a^2+b^2+c^2+d^2+w^2+x^2+y^2+z^2=n$$
We are interested in interpreting $\sum_{n=1}^x r_8(n)$. One way is to think of it as an approximation of the $8$-dimensional ball which has volume $\frac{\pi^4}{24}x^4$ and the other is to exploit a formula for $r_8(n)$ to examine the asymptotic behavior of $\sum_{n=1}^x r_8(n)$.
We will need that $\sum_{n=1}^x \sigma_k(x) \approx \frac{\zeta(k+1)} {k+1}x^{k+1}$ to make this precise we may say that $$\lim_{x\to\infty}{\frac{k+1}{x^{k+1}}\sum_{n=1}^x{\sigma_k(n)}}=\zeta(k+1)$$
I think that this can be seen with just a slight rewriting of this. 
From equation (12) of this paper we can write 
$r_8(n)=16\sigma_3(n)-32\sigma_3(n/2)+256\sigma_3(n/4)$ 
Now equipped with $\sum_{n=1}^x\sigma_3\big(\frac{n}{a}\big)\approx \frac{\zeta(4)}{4}\big(\frac{x}{a}\big)^4$
We may write that 
$$\begin{align}
&\frac{\pi^4}{24}x^4 \\
&\approx&\sum_{n=1}^x r_8(n) \\ 
&=&16\sum_{n=1}^x {\sigma_3(n)}
&-32\sum_{n=1}^x {\sigma_3\bigg(\frac{n}{2}\bigg)}
&+256\sum_{n=1}^x {\sigma_3\bigg(\frac{n}{4}\bigg)}\\
&\approx &16 \frac{\zeta(4)}{4}x^4
&-32\frac{\zeta(4)}{4}\bigg(\frac{x}{2}\bigg)^4 
&+256 \frac{\zeta(4)}{4}\bigg(\frac{x}{4}\bigg)^4 
\\
&=x^4\zeta(4)\frac{15}{4}
\end{align}$$
So we have that as we let $x$ grow large 
$$\frac{\pi^4}{24}x^4 \approx \frac{15}{4}\zeta(4)x^4$$ 
So then we can conclude that $\zeta(4)=\pi^4/90$
