special values of symmetric functions at powers of $\frac1j$ Let $e_n(x_1,x_2,x_3,\dots)$ denote the $n$-th elementary symmetric function in the infinite variables $x_1,x_2,x_3,\dots$.
Let $u$ and $v$ be the roots of $z^2-6z+1=0$.

Question. Let $x_j=\frac1{j^8}$. The following seems to be true, but can one prove or disprove?
$$e_n(x_1,x_2,x_3,\dots)=\frac{4^{3n+1}\pi^{8n}(u^{2n+1}+v^{2n+1})}{(8n+4)!}.$$

For example, $e_0=1$ and $e_1=\sum_{j\geq1}\frac1{j^8}=\zeta(8)=\frac{\pi^8}{9450}$.
 A: As Gro-Tsen suggests in the comments, we have to expand the infinite product $$f(t)=\prod_j \left(1-\frac{t^8}{j^8}\right)=\prod_{j;\,w^4=1} \left(1-\frac{\omega t^2}{j^2}\right)=\prod_{w^4=1}\frac{\sin\pi\sqrt{w}t}{\pi\sqrt{w}t},$$
we expand product of four sines as an alternating sum of cosines $$\sin a\sin b\sin c\sin d=\frac18\sum (\prod \pm)\cdot\cos(a\pm b\pm c\pm d),$$
and write Taylor series for cosines. Powers of $1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)$ appear. To be more concrete, $c_n:=e_n(x_1,\dots)$ equals $(-1)^n\times [t^{8n}]f(t)$, where $[t^n]g$ denotes a coefficient of $t^n$ in $g$. Thus $$c_n=(-1)^{n+1}i\pi^{8n}\frac1{8(8n+4)!}\sum \pm\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^{8n+4}.$$
We have $$\left(1\pm i\pm(\frac{\sqrt{2}}2+i\frac{\sqrt{2}}2)\pm (\frac{\sqrt{2}}2-i\frac{\sqrt{2}}2)\right)^4=8i(\pm 3\pm 2\sqrt{2}),$$
that gives your formula after careful simplifications.
