Euler representation of Gamma function proof

$$\Gamma(z) = \lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = \frac{1}{z}\prod^\infty_{k=1}\frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}} $$

$$  \textit{proof} $$

Note that

$$\Gamma(z+n+1) = \Gamma(z+1) \prod^{n}_{k=1}(k+z)$$

Which indicates that

$$\prod^{n}_{k=1}(k+z) = \frac{\Gamma(z+n+1)}{z\Gamma(z)}$$

Also note that

$$\prod^{n}_{k=1}k = n!$$

Hence we have

$$\lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} = \Gamma(z)\lim_{n \to \infty }\frac{n^z\times n!}{ \Gamma(z+n+1)}$$

Hence we must show that

$$\lim_{n \to \infty }\frac{n^z\times n!}{\Gamma(z+n+1)} = 1$$

Note that by Stirling formula

$$\Gamma(z+n+1) \sim \sqrt{2\pi} (n+z)^{n+z+1/2} e^{-(n+z)}$$

and

$$n! \sim \sqrt{2\pi} n^{n+1/2}e^{-n}$$

Hence we have by

$$ \lim_{n \to \infty }\frac{n^z\times (\sqrt{2\pi} n^{n+1/2}e^{-n})}{\sqrt{2\pi} (n+z)^{n+z+1/2} e^{-(n+z)}} = \lim_{n \to \infty }\frac{n^{n}}{(n+z)^{n} e^{-z}} = 1 $$

Note that $$\lim_{n \to \infty }\left(1+\frac{z}{n} \right)^n = e^{z}$$

To prove the other product formula note that

$$\prod^n_{k=1}\left( 1+\frac{1}{k}\right)^z = \frac{\prod^n_{k=1}\left( 1+k\right)^z}{\prod^n_{k=1} k^z} = (n+1)^z \sim n^z $$

Hence we deduce

$$\lim_{n \to \infty }\frac{n^z}{z}\prod^n_{k=1} \frac{k}{k+z} =\lim_{n \to \infty }\frac{\prod^n_{k=1} \left(1+\frac{1}{k}\right)^z}{z}\prod^n_{k=1} \frac{1}{1+\frac{z}{k}} = \frac{1}{z}\prod^\infty_{k=1}\frac{\left(1+\frac{1}{k}\right)^z}{1+\frac{z}{k}} $$

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