Sometime back I made a claim here that the proof for $\zeta(4)$ can be extended to all even numbers.
I tried doing this but I face a stumbling block.
Let me explain the problem in detail here. I was trying to mimic Euler's proof for the Basel problem to evaluate $\zeta$ at all even integers and prove that $\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $
Euler looks at the function whose zeros are at $\pm \pi, \pm 2 \pi, \pm 3 \pi, \ldots$.
In general, to evaluate $\zeta(2n)$, let us look at a function whose roots are $\pm \xi_0 \pi, \pm \xi_1 \pi, \ldots \pm \xi_{n-1} \pi; \pm \xi_0 2\pi, \pm \xi_1 2\pi, \ldots, \pm \xi_{n-1} 2\pi; \pm \xi_0 3\pi, \pm \xi_1 3\pi, \ldots, \pm \xi_{n-1} 3\pi; \ldots$ where $\xi_k = \exp \left( \dfrac{k \pi i}{n} \right)$, $k \in \{0,1,2,\ldots,n-1\}$.
Let $p_{2n}(z) = \left(1 - \left(\frac{z}{\pi}\right)^{2n} \right) \times \left(1 - \left(\frac{z}{2 \pi}\right)^{2n} \right) \times \left(1 - \left(\frac{z}{3 \pi}\right)^{2n} \right) \times \cdots$
The coefficient of $z^{2n}$ in $p_{2n}(z)$ is $- \dfrac1{\pi^{2n}}\sum_{k=1}^{\infty} \dfrac1{k^{2n}} = - \dfrac{\zeta(2n)}{\pi^{2n}}$
It is not hard to guess that $p_{2n}(z)$ is same as $\prod_{k=0}^{n-1} \dfrac{\sin(z/\xi_k)}{(z/\xi_k)} = \exp \left( \dfrac{(n-1) \pi i}2\right) \left(\dfrac{\sin(z/\xi_0) \sin(z/\xi_1) \sin(z/\xi_2) \ldots \sin(z/\xi_{n-1})}{z^{n}} \right)$
Now from the power series expansion of $\sin(z)$, we get that $\dfrac{\sin(z/\xi_k)}{z/\xi_k} = \left(\sum_{\ell=0}^{\infty} \dfrac{(-1)^{\ell} z^{2 \ell}}{(2 \ell+1)! \xi_k^{2\ell}} \right)$
Hence, we get that $\prod_{k=0}^{n-1} \dfrac{\sin(z/\xi_k)}{(z/\xi_k)} = \prod_{k=0}^{n-1} \left(\sum_{\ell=0}^{\infty} \dfrac{(-1)^{\ell} z^{2 \ell}}{(2 \ell+1)! \xi_k^{2\ell}} \right)$
The coefficient of $z^{2n}$ in the product is given by $c(n) = \sum_{\overset{\ell_0 + \ell_1 + \cdots + \ell_{n-1} = n}{\ell_k \geq 0}} \prod_{k=0}^{n-1} \left( \dfrac{(-1)^{\ell_k}}{(2 \ell_k+1)! \xi_k^{2 \ell_k}} \right)$
Assuming whatever I have done so far is correct, to complete the proof and conclude that $\zeta(2n) = (-1)^{n+1} \frac{B_{2n} 2^{2n}}{2(2n)!} \pi^{2n} $ I need to prove the following claim
Claim: $\color{red}{B_{2n} = \dfrac{(2n)!}{2^{2n-1}}\sum_{\overset{\ell_0 + \ell_1 + \cdots + \ell_{n-1} = n}{\ell_k \geq 0}} \prod_{k=0}^{n-1} \left( \dfrac{1}{(2 \ell_k+1)! \xi_k^{2 \ell_k}} \right)}$ where $\xi_k = \exp \left( \dfrac{k \pi i}{n} \right)$.
Is the above expression of the Bernoulli numbers a known result? If so, could someone point me to a proof of this result?