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

Let $ f(x)=x^2-\frac{1}{12} $ defined on $ [-\frac{1}{2},-\frac{1}{2}) $, and extend $ f $ to be periodic on $ \mathbb{R} $.
We compute that 
    \begin{align*}
 \hat f(0)=\int_{-1/2}^{1/2}f(x)dx=\int_{-1/2}^{1/2}\left(x^2-\frac{1}{12}\right)dx=0.
 \end{align*}
    And for $ k\ne 0 $:
    \begin{align*}
 \hat f(k)&=\int_{-1/2}^{1/2}f(x)e^{-2\pi ikx}dx=\int_{-1/2}^{1/2}\left(x^2-\frac{1}{12}\right)e^{-2\pi ikx}dx=\int_{-1/2}^{1/2}x^2e^{-2\pi ikx}dx\\
 &=-\left.\frac{x^2e^{-2\pi i kx}}{2\pi ik}\right|_{-1/2}^{1/2}+\frac{1}{\pi ik}\int_{-1/2}^{1/2}xe^{-2\pi ikx}dx=\frac{1}{\pi ik}\int_{-1/2}^{1/2}xe^{-2\pi ikx}dx\\
 &=\frac{1}{2\pi^2k^2 }\int_{-1/2}^{1/2}xd(e^{-2\pi ikx})=\left.\frac{1}{2\pi^2k^2}xe^{-2\pi ikx}\right|_{-1/2}^{1/2}+\frac{1}{2\pi^2k^2}\int_{-1/2}^{1/2} e^{-2\pi ikx}dx=\frac{(-1)^k}{2\pi^2k^2}.
 \end{align*}
By the Parseval identity
    \begin{align*}
 \int_{-1/2}^{1/2}|f(x)|^2dx=\sum_{k=-\infty}^{\infty}|\hat{f}(k)|^2=|\hat{f}(0)|^2+2\sum_{k=1}^{\infty}|\hat{f}(k)|^2=\sum_{k=1}^{\infty}\frac{1}{2\pi^4 k^4}. 
 \end{align*}
    On the other hand,
    \begin{align*}
 \int_{-1/2}^{1/2}|f(x)|^2dx&=\frac 1{180}.
 \end{align*}
    Hence, we have $$ \sum_{k=1}^{\infty}\frac{1}{k^4}=\frac{\pi^4}{90}. $$
