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Question Number 117816 by snipers237 last updated on 13/Oct/20

$$findoutforn⩾1$$$$\underset{k=0}{\prod ^{n-1}}\mathrm{\Gamma }(1+\frac{k}{n})$$$$$$

Answered by mnjuly1970 last updated on 14/Oct/20

$$A=\mathrm{\Gamma }\left(1\right)\mathrm{\Gamma }(1+\frac{1}{n})\mathrm{\Gamma }(1+\frac{2}{n})..\mathrm{\Gamma }(1+\frac{n-1}{n})$$$$A^2={(\frac{1}{n}\ast \frac{2}{n}\ast \frac{3}{n}...\ast \frac{n-1}{n})}^2\ast (\frac{\pi }{sin\left(\frac{\pi }{n}\right)}\ast \frac{\pi }{sin\left(\frac{2\pi }{n}\right)}...\ast \frac{\pi }{sin\left(\frac{(n-1)\pi }{n}\right)}$$$$$$$$=\frac{{(n!)}^2}{n^{2(n-1)}}\ast \frac{{\pi }^{n-1}\ast 2^{n-1}}{n}$$$$=\frac{{(n!)}^{2}n{\left(2\pi \right)}^{n-1}}{n^{2n}}$$$$A=\frac{n!\sqrt{n{\left(2\pi \right)}^{n-1}}}{n^n}✓$$$$m.n.1970$$$$$$

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