How can one prove that
$$\zeta (2n)=\frac{(-1)^{n-1}2^{2n-1}\pi ^{2n}B_{2n}}{(2n)!}$$ where $n\in N$
and how can one prove that $$\zeta (2n)=\frac{(-1)^{n}2^{2n-2}\pi ^{2n}E_{2n-1}}{(2n-1)!(2^{2n}-1)}$$ where $n\in N^*$
and $\zeta (x)$ is the Riemann zeta function, $B_n$ a Bernoulli number, and $E_n$ an Euler number?
edit 2 after the comments: is there any link or book can give answer to my Question ?
In the product representation $$\sin z = z \prod_{k=1}^{\infty} \left(1-\frac{z^2}{(k\pi)^2}\right)$$, we obtain $$\log(\sin z) = \log z + \sum_{k=1}^{\infty} \log \left(1-\frac{z^2}{(k\pi)^2}\right)$$.
When differentiated, $$\cot z = \frac{1}{z} - \frac{2}{z} \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \left(\frac{z}{k\pi}\right)^{2n} = \frac{1}{z} + \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{-2}{(k\pi)^{2n}} \, z^{2n-1}$$
By comparing coefficients with $$\cot z = \frac{1}{z} + \sum_{n=1}^{\infty} (-1)^{n} \frac{B_{2n} \, 2^{2n}}{(2n)!} \, z^{2n-1}$$, we obtain $$\sum_{k=1}^{\infty} \frac{1}{(k\pi)^{2n}} = (-1)^{n-1} \frac{B_{2n} \, 2^{2n-1}}{(2n)!}$$
