proof of multiplication formula for gamma function

Define the function $f$ as

$$f(z)=\frac{n^{nz}\prod _{k=0}^{n-1}\mathrm{\Gamma }\left(z+\frac{k}{n}\right)}{\mathrm{\Gamma }(nz)}$$

By the functional equation of the gamma function,

$$f(z+1)=\frac{n^n{n}^{nz}\left(\prod _{m=0}^{n-1}\mathrm{\Gamma }\left(z+\frac{m}{n}\right)\right)\prod _{k=0}^{n-1}\left(z+\frac{k}{n}\right)}{\mathrm{\Gamma }(nz){\prod }_{k=0}^{n-1}(nz+k)}=f(z)$$

Hence $f$ is a periodic function of $z$. However, for large valuesof $z$, we can apply the Stirling approximation formula to conclude

$$f(z)=\frac{{(2\pi )}^{n/2}{n}^{nz}\prod _{k=0}^{n-1}\left[e^{-z-k/n}{(z+k/n)}^{z+k/n-1/2}+O(e^{-z}{(z+k/n)}^{z+k/n-3/2})\right]}{{(2\pi )}^{1/2}{e}^{-nz}{(nz)}^{nz-1/2}+O(e^{-nz}{(nz)}^{nz-3/2})}=$$

$${(2\pi )}^{(n-1)/2}{n}^{1/2}\frac{\prod _{k=0}^{n-1}\left[e^{-k/n}{(z+k/n)}^{z+k/n-1/2}+O({(z+k/n)}^{z+k/n-3/2})\right]}{z^{nz-1/2}+O(z^{nz-3/2})}=$$

$${(2\pi )}^{(n-1)/2}{n}^{1/2}\frac{z^{1/2}\prod _{k=0}^{n-1}\left[e^{-k/n}{\left(1+\frac{k}{nz}\right)}^{z+k/n-1/2}{z}^{k/n-1/2}+O({(z+k/n)}^{k/n-3/2})\right]}{1+O(z^{-1})}$$

Note that

$$\prod _{k=0}^{n-1}{e}^{-k/n}=e^{-{\sum }_{k=0}^{n-1}k/n}=e^{(1-n)/2}$$

$$z^{1/2}\prod _{k=0}^{n-1}{z}^{k/n-1/2}=z^{1/2}{z}^{{\sum }_{k=0}^{n-1}(k/n-1/2)}=z^{1/2+(n-1)/2-n/2}=1$$

Also,

$${\left(1+\frac{k}{nz}\right)}^z=e^{k/n}+O(z^{-1})$$

Hence, $f(z)={(2\pi )}^{(n-1)/2}{n}^{1/2}+O(z^{-1})$. Now, the only way for a function to be periodic and have a definite limit is for that function to be constant. Therefore, $f(z)={(2\pi )}^{(n-1)/2}{n}^{1/2}$. Writing out the definition of $f$ and rearranging gives the multiplication formula.