ガンマ関数の倍数公式のスターリングの公式を用いない証明

$\begin{array}{rl}(\text{左}\text{辺})& =\prod _{k=1}^{n-1}\left(i\sinh (-\frac{k\pi i}{n})\right)={\left(\frac{i}{2}\right)}^{n-1}\prod _{k=1}^{n-1}(e^{-\frac{k\pi i}{n}}-e^{\frac{k\pi i}{n}})\\ & ={\left(\frac{i}{2}\right)}^{n-1}{e}^{-\frac{\pi i}{n}\frac{n(n-1)}{2}}\prod _{k=1}^{n-1}(1-e^{\frac{2k\pi i}{n}})\\ & ={\left(\frac{i}{2}\right)}^{n-1}{e}^{-\frac{(n-1)\pi i}{2}}\underset{a\to 1}{lim}\frac{1}{a-1}\prod _{k=0}^{n-1}(a-e^{\frac{2k\pi i}{n}})\\ & ={\left(\frac{i}{2}\right)}^{n-1}{i}^{-n+1}\underset{a\to 1}{lim}\frac{a^n-1}{a-1}=(\text{右}\text{辺})\end{array}$

補題3を両辺積分して、
$\begin{array}{rl}\ln \mathrm{\Gamma }(nz)& =n\ln n\cdot z+\sum _{k=0}^{n-1}\ln \mathrm{\Gamma }(z+\frac{k}{n})+C\\ \mathrm{\Gamma }(nz)& =D{n}^{nz}\prod _{k=0}^{n-1}\mathrm{\Gamma }(z+\frac{k}{n})\left(D:=e^C\right)\\ & z=\frac{1}{n}\mathsf{\text{を代入して、}}\\ D& =\frac{\mathrm{\Gamma }\left(\frac{n}{n}\right)}{n\prod _{k=0}^{n-1}\mathrm{\Gamma }(\frac{1}{n}+\frac{k}{n})}=\frac{1}{n\prod _{k=1}^{n-1}\mathrm{\Gamma }\left(\frac{k}{n}\right)}\\ & =\frac{1}{n\sqrt{\prod _{k=1}^{n-1}\mathrm{\Gamma }\left(\frac{k}{n}\right)\mathrm{\Gamma }(1-\frac{k}{n})}}\\ & =\frac{1}{n}\sqrt{\prod _{k=1}^{n-1}\frac{\sin \frac{k\pi }{n}}{\pi }}& & \because \mathsf{\text{相反公式}}\\ & =\frac{\sqrt{{\pi }^{1-n}{2}^{1-n}n}}{n}& & \because \mathsf{\text{補題2}}\\ & =\frac{(2\pi {)}^{\frac{1-n}{2}}}{\sqrt{n}}\\ & \mathsf{\text{よって、}}\\ \mathrm{\Gamma }(nz)& =\frac{(2\pi {)}^{\frac{1-n}{2}}}{\sqrt{n}}{n}^{nz}\prod _{k=0}^{n-1}\mathrm{\Gamma }(z+\frac{k}{n})\\ & =(2\pi {)}^{\frac{1-n}{2}}{n}^{nz-\frac{1}{2}}\prod _{k=0}^{n-1}\mathrm{\Gamma }(z+\frac{k}{n})\end{array}$