Gauss multiplication formula

Remarks

For $x > 0$, the function $f(x) = \frac{m^{x - 1/2}}{(2\pi)^{(m-1)/2}} \prod_{k=0}^{m-1} \Gamma\!\left(\frac{x + k}{m}\right)$ satisfies $f(x+1) = x f(x)$ and $\ln f$ is convex. So $f = c\,\Gamma$ for a constant $c$, by the argument of the Bohr–Mollerup theorem; the constant is found from $f(1) = 1$, which uses $\prod_{k=1}^{m-1}\Gamma(k/m) = (2\pi)^{(m-1)/2}/\sqrt{m}$, a consequence of the reflection formula. Alternatively, the constant follows from Stirling's formula as $z \to \infty$. For $m = 1$ the formula is trivial, and for $m = 2$ it is the Legendre duplication formula, since $(2\pi)^{1/2} 2^{1/2 - 2z} = 2^{1-2z}\sqrt{\pi}$.

Going deeper

For a student

Taking logarithms and differentiating gives the matching multiplication formula for the digamma function. Similar multiplication theorems hold for other special functions, such as the Hurwitz zeta function and the Bernoulli polynomials. Setting $z = 1/m$ gives the product $\prod_{k=1}^{m-1}\Gamma(k/m) = (2\pi)^{(m-1)/2}/\sqrt{m}$.