ガンマ関数
両辺の比を f ( z ) {\displaystyle f(z)} とすると
f ( z ) = n n z − 1 / 2 ∏ k = 0 n − 1 Γ ( z + k n ) ( 2 π ) ( n − 1 ) / 2 Γ ( n z ) {\displaystyle {\begin{aligned}f(z)=&{\frac {n^{nz-1/2}\prod _{k=0}^{n-1}{\Gamma {\left(z+{\frac {k}{n}}\right)}}}{(2\pi )^{(n-1)/2}\Gamma (nz)}}\\\end{aligned}}} f ( z + 1 ) = n n z − 1 / 2 n n [ ∏ k = 0 n − 1 ( z + k n ) Γ ( z + k n ) ] ( 2 π ) ( n − 1 ) / 2 [ ∏ k = 0 n − 1 ( n z + k ) ] Γ ( n z ) = n n z − 1 / 2 [ ∏ k = 0 n − 1 ( n z + k ) ] ∏ k = 0 n − 1 Γ ( z + k n ) ( 2 π ) ( n − 1 ) / 2 [ ∏ k = 0 n − 1 ( n z + k ) ] Γ ( n z ) = f ( z ) {\displaystyle {\begin{aligned}f(z+1)&={\frac {n^{nz-1/2}n^{n}\left[\prod _{k=0}^{n-1}\left(z+{\frac {k}{n}}\right)\Gamma {\left(z+{\frac {k}{n}}\right)}\right]}{(2\pi )^{(n-1)/2}\left[\prod _{k=0}^{n-1}(nz+k)\right]\Gamma (nz)}}\\&={\frac {n^{nz-1/2}\left[\prod _{k=0}^{n-1}\left(nz+k\right)\right]\prod _{k=0}^{n-1}\Gamma {\left(z+{\frac {k}{n}}\right)}}{(2\pi )^{(n-1)/2}\left[\prod _{k=0}^{n-1}(nz+k)\right]\Gamma (nz)}}\\&=f(z)\\\end{aligned}}}