In this paper, Yaglom shows an elementary derivation of the solution to the Basel problem by considering polynomials of the form $$\sum_{k=0}^m(-1)^k\binom{2m+1}{2k+1}x^{m-k}$$ which has roots $$\displaystyle \cot^2\frac{k\pi}{2m+1},\,k=1,2,\ldots,m$$ For convenience of notation let $n=2m+1$. Using Vieta's relations, Yaglom shows that $$\sum_{k=1}^m\cot^2\frac{k\pi}{n}=\frac{\binom{n}{3}}{\binom{n}{1}}=\frac{(n-1)(n-2)}{6}$$ Using the trig identity $1+\cot^2\theta=\csc^2\theta$, we get $$\sum_{k=1}^m\csc^2\frac{k\pi}{n}=\frac{(n-1)(n-2)}{6}+\frac{n-1}{2}=\frac{(n-1)(n+1)}{6}$$ Finally, using $\displaystyle\cot^2\theta<\frac{1}{\theta^2}<\csc^2\theta\,$ for $\theta\in (0,\pi/2)$, we get $$\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\frac{\pi^2}{6}<\sum_{k=1}^m\frac{1}{k^2}<\left(1-\frac{1}{n}\right)\left(1+\frac{1}{n}\right)\frac{\pi^2}{6}$$ which gives us the desired solution. Yaglom suggests that this method can be extended via Newton's identities to produce values for the Riemann Zeta function on even natural numbers. If we call $$p_j=\sum_{k=1}^m\cot^{2j}\frac{k\pi}{n}$$ then by Newton's identities we get the recurrence relation $$k\binom{n}{2k+1}=\sum_{j=1}^k(-1)^{j-1}\binom{n}{2(k-j)+1}p_j$$ Unfortunately I don't have much experience with recurrence relations, and I have no idea how to proceed. Any help would be very much appreciated. My goal is to find and alternate expression for $p_j$, just as Yaglom found that $$p_1=\frac{\binom{n}{3}}{\binom{n}{1}}$$
Edit: I'm specifically looking for a way to derive $$\zeta(2k)=\frac{(-1)^{k+1}B_{2k}(2\pi)^{2k}}{2(2k)!}$$ using this method.