$-2(-1)^n\zeta(n)=\lim_{m\to\infty}m^n\left(\displaystyle\prod_{k=1}^n\Gamma\left(1+\frac{\zeta_n^k}{m}\right)^{-2}-1\right).$ 
Theorem
Let $\zeta_n=e^{2\pi i/n}$ be the $n$-th root of unity, then
$$-2(-1)^n\zeta(n)=\lim_{m\to\infty}m^n\left(\displaystyle\prod_{k=1}^n\Gamma\left(1+\frac{\zeta_n^k}{m}\right)^{-2}-1\right).$$

Just wanted to share this - I thought this up a couple of years ago, but have no proof - it was a side thought. I'll post a proof when I get a moment, but if anyone can see how to do it that would be great :-)

The series $\log\Gamma(1-z)=\gamma z+\sum_{m=2}^\infty\zeta(m)z^m/m$ (convergent for $|z|<1$) can be obtained from the Weierstrass product for $\Gamma$. Put $z=-t\zeta_n^k$ where $t\to0$ and $\color{red}{n>1}$, and sum over $k$: $$\sum_{k=1}^n\zeta_n^{km}=\begin{cases}n,&n\mid m\\0,&n\nmid m\end{cases}\implies\sum_{k=1}^n\log\Gamma(1+t\zeta_n^k)\underset{m=nk}{=}\sum_{k=1}^\infty\zeta(nk)(-t)^{nk}/k.$$ Thus $\prod_{k=1}^n\Gamma(1+t\zeta_n^k)^{-2}=\exp\big(-2\zeta(n)(-t)^n+o(t^n)\big)=1+2(-1)^{n-1}\zeta(n)t^n+o(t^n)$.
