Let φ(n) denote Euler's totient function and k $\perp$ n denote that k, n are integers and relatively prime. Let N = φ(n) + 1. If n is not a prime power
$$ \prod\_{\substack{0 < k < n, \\ k \perp n }} \Gamma \left(\frac{k}{n}\right)
 = \sqrt{N}\prod\_{ 0 < k < N}\Gamma \left(\frac{k}{N}\right) \quad (n \neq p^a) . $$
This identity was proved [here](http://arxiv.org/abs/0909.1838) as a corollary to some other identities, but the authors asked: ``is there some natural direct proof of this formula?´´ 



A product of gamma values over the numbers coprime to n.
Denote $$f(n)=\prod\_{k=1}^{n-1}\Gamma \left(\frac{k}{n}\right)$$ and $$ F(n)=\prod\_{1\le k\le n-1, k\perp n}\Gamma \left(\frac{k}{n}\right)$$
We have $f(n)=\prod\_{d|n}F(n)$ and therefore by Mobius inversion $F(n)=\prod\_{d|n}f(d)^{\mu(n/d)}$


By the multiplication theorem we have $f(n)=\frac{1}{\sqrt{n}}(2\pi)^{\frac{n-1}{2}}$, so if $n$ is not a prime power $$F(n)=\prod\_{d|n}\left(\frac{1}{\sqrt{d}}(2\pi)^{\frac{d-1}{2}}\right)^{\mu(n/d)}=(2\pi)^{\frac{1}{2}\varphi (n)}$$
The formula $F(n)=\sqrt{\varphi(n)+1}f(\varphi(n)+1)$ follows.


