A Bernoulli number identity and evaluating $\zeta$ at even integers 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?
 A: As far as I can tell, you have already proven your claim; you started with a function with the appropriate value for the coefficient of interest and then found an expression for that coefficient. Consequently that expression must be a Bernoulli number by construction. If for some reason you are after an alternative proof that the coefficient of $z^{2n}$ in the function:
$g(z)=2(-1)^n(2n)!2^{-2n}f(z)$, 
where:
$f \left( z \right) =\prod _{k=0}^{n-1}\mathrm{sinc} \left( z \exp \left( -\dfrac{k \pi i}{n} \right)\right) $, 
$\mathrm{sinc}(x)=\dfrac{\sin(x)}{x},$
is equal to the Bernoulli number $B(2n)$ then here's one... 
First take the logarithm of $f(z)$:
$\ln(f \left( z \right)) =\sum _{k=0}^{n-1}\ln\left(\mathrm{sinc} \left( z \exp \left(- \dfrac{k \pi i}{n} \right)\right)\right)$. 
Then use the product expansion of the sinc function:
$\mathrm{sinc} \left( z \exp \left( -\dfrac{k \pi i}{n} \right)\right)=\prod _{q=1}^{\infty}\left(1-\left(\dfrac{z}{{\pi}q}\right)^2\exp \left(- \dfrac{2k \pi i}{n} \right)\right)$ 
to obtain: 
$\ln(f \left( z \right)) =\sum _{k=0}^{n-1}\sum _{q=1}^{\infty}\ln\left(1-\left(\dfrac{z}{{\pi}q}\right)^2\exp \left(- \dfrac{2k \pi i}{n} \right)\right)$. 
then the series expansion of the logarithm: 
$\ln\left(1-\left(\dfrac{z}{{\pi}q}\right)^2\exp \left(- \dfrac{2k \pi i}{n} \right)\right)=-\sum _{r=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2r}r^{-1} \exp \left(- \dfrac{2kr \pi i}{n} \right)$ 
to get: 
$\ln(f \left( z \right)) =-\sum _{k=0}^{n-1}\sum _{q=1}^{\infty}\sum _{r=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2r}r^{-1} \exp \left(- \dfrac{2kr \pi i}{n} \right)$. 
$\ln(f \left( z \right)) =-\sum _{q=1}^{\infty}\sum _{r=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2r}r^{-1} \sum _{k=0}^{n-1}\exp \left(- \dfrac{2kr \pi i}{n} \right)$. 
Note the Kronecker delta popping up that will only pick out $r$ when it's a multiple (denoted $l$) of $n$:
$\sum _{k=0}^{n-1}\exp \left(- \dfrac{2kr \pi i}{n} \right)=n\delta_{r,nl}$, 
so we may multiply by $n$, replace $r$ with $nl$ and sum over $l$:
$\ln(f \left( z \right)) =-\sum _{q=1}^{\infty}\sum _{l=1}^{\infty}\left(\dfrac{z}{{\pi}q}\right)^{2nl}l^{-1} $. 
Now recognise the sum over $q$:
$\sum _{q=1}^{\infty}\dfrac{1}{q^{2nl}}=\zeta(2nl)$
and so it follows:
$\ln(f \left( z \right)) =-\sum _{l=1}^{\infty}\zeta(2nl)\left(\dfrac{z}{{\pi}}\right)^{2nl}l^{-1} $. 
Now take the exponential:
$f \left( z \right) =\exp\left(-\sum _{l=1}^{\infty}\zeta(2nl)z^{2nl}{\pi}^{-2nl}l^{-1} \right)$, 
$f \left( z \right) =\prod _{l=1}^{\infty}\exp\left(-\zeta(2nl)z^{2nl}{\pi}^{-2nl}l^{-1} \right)$, 
$f \left( z \right) =\prod _{l=1}^{\infty}\sum _{u=0}^{\infty}\left(-\zeta(2nl)z^{2nl}{\pi}^{-2nl}l^{-1} \right)^u(u!)^{-1}$. 
We are interested in the coefficient of $z^{2n}$ and this will not receive any contribution from $1<l$ or $1<u$ and thus: 
$f \left( z \right) =1-\zeta(2n)z^{2n}{\pi}^{-2n}+O(z^{4n})$, 
and consequently:
$g(z)=2(-1)^n(2n)!2^{-2n}f(z)= 2(-1)^n(2n)!2^{-2n}-2(-1)^n(2n)!2^{-2n}\zeta(2n)z^{2n}{\pi}^{-2n}+O(z^{4n})$. 
From which it follows that the coefficient of $z^{2n}$ is:
$-2(-1)^n(2n)!\zeta(2n)(2\pi)^{-2n}=B(2n)$.
As you have supplied an alternative expression for the same coefficient your expression must also equal $B(2n)$.
It seems like an interesting expression, do you have an application for it? Can you use it to calculate Fourier coefficients in products of series for example or something like that?
