I have posted here in Portuguese a recursive method based on the computation of the Fourier trigonometric series expansion for the function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{2p}$ and extended to all of ${\mathbb R}$ periodically with period $2\pi.$ This is a shorter description than the original. In this reply I outline the case $\zeta(4)$. For $p=3$ the expansion is
$x^{6}=\dfrac{\pi ^{6}}{7}+2\displaystyle\sum_{n\ge 1}^{}\left( \left( \dfrac{6}{n^{2}}\pi ^{4}-\dfrac{120}{n^{4}}\pi ^{2}+\dfrac{720 }{n^{6}}\right)\cos n\pi \right) \cos nx.\tag{1}$
The computation is as follows:
$\begin{equation*} f(x)=x^{2p}=\frac{a_{0,2p}}{2}+\sum_{n=1}^{\infty }\left( a_{n,2p}\cos nx+b_{n,2p}\sin nx\right) , \end{equation*}$ where the coefficients are given by the following integrals $\begin{eqnarray*} a_{0,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\;\mathrm{d}x=\frac{2\pi ^{2p}}{2p+1}, \\ a_{n,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\cos nx\;\mathrm{d}x=\frac{2}{\pi } \int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x, \\ b_{n,2p} &=&\frac{1}{\pi }\int_{-\pi }^{\pi }x^{2p}\sin nx\;\mathrm{d}x=0. \end{eqnarray*}$ The series expansion is thus $\begin{equation*} x^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos nx\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x.\tag{2} \end{equation*}$ For $f(\pi )=\pi ^{2p}$ we obtain $ \begin{equation*} \pi ^{2p}=\frac{\pi ^{2p}}{2p+1}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi \int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x, \end{equation*}$ where the integral $ \begin{equation*} I_{n,2p}:=\int_{0}^{\pi }x^{2p}\cos nx\;\mathrm{d}x \end{equation*}$ satisfies the following recurrence, as can be shown by integration by parts $\begin{equation*} I_{n,2p}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi -\frac{2p(2p-1)}{n^{2}} I_{n,2\left( p-1\right) },\qquad I_{n,0}=0.\tag{3} \end{equation*}$
- For $p=1$, we get $\begin{equation*} I_{n,2}=\frac{2p}{n^{2}}\pi ^{2p-1}\cos n\pi. \end{equation*}$ and $\begin{eqnarray*} \pi ^{2} &=&\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi \cdot I_{n,2} \\ &=&\frac{\pi ^{2}}{3}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi \left( \frac{2}{n^{2}}\pi \cos n\pi \right) \\ &=&\frac{\pi ^{2}}{3}+4\sum_{n=1}^{\infty }\frac{1}{n^{2}} \\ &\Rightarrow &\zeta (2)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}=\frac{\pi ^{2}}{6 } \end{eqnarray*}$
- For $p=2$, we get $ \begin{equation*} I_{n,4}=\left( \frac{4\pi ^{3}}{n^{2}}-\frac{24\pi }{n^{4}}\right) \cos n\pi \end{equation*}$ and $ \begin{eqnarray*} \pi ^{4} &=&\frac{\pi ^{4}}{5}+\frac{2}{\pi }\sum_{n=1}^{\infty }\cos n\pi \cdot I_{n,4}=\frac{\pi ^{4}}{5}+\frac{4\pi ^{4}}{3}-48\sum_{n=1}^{\infty } \frac{1}{n^{4}} \\ &\Rightarrow &\zeta (4)=\sum_{n=1}^{\infty }\frac{1}{n^{4}}=\frac{\pi ^{4}}{ 90}. \end{eqnarray*}$
- Finally for $p=3$, we get $\begin{equation*} I_{n,6}=\left( \frac{6\pi ^{5}}{n^{2}}-\frac{120\pi ^{3}}{n^{4}}+\frac{720}{ n^{6}}\right) \cos n\pi \end{equation*}$ and $ \begin{equation*} \pi ^{6}=\frac{\pi ^{6}}{7}+2\sum_{n=1}^{\infty }\left( \frac{6\pi ^{4}}{ n^{2}}-\frac{120\pi ^{2}}{n^{4}}+\frac{720}{n^{6}}\right), \end{equation*}$ from which the result follows $\zeta(6)= \begin{equation*} \sum_{n=1}^{\infty }\frac{1}{n^{6}}=\frac{\pi ^{6}}{945}. \end{equation*}$
Plots of the periodic function defined in $\left[ -\pi ,\pi \right] $ by $f(x)=x^{6}$ (blue curve) and of the partial sum with the first 10 terms of its Fourier trigonometric series (red curve).

This method generates recursively the sequence $(\zeta(2p))_{p\ge 1}$.