Prove that Gamma(x) = x Gamma(x)

I will here go over both the proofs

$\Gamma(x+1) = x\Gamma(x)$

Integration method: Use integration by parts on

$\Gamma(x+1) = \int_0^{\infty} t^{x}e^{-t} dt$

$\Gamma(x+1) = x\int_0^{\infty} t^{x-1} e^{-t} dt$

$\Gamma(x+1) = x \Gamma(x)$

The 2nd method is using the Euler’s definition

$\Gamma(x) = \lim_{ r \to \infty } \Gamma_r(x) = \lim_{r \to \infty} \dfrac{r! r^x}{x(x+1)...(x+r)}$

$\Gamma(x+1) = \lim_{r \to \infty} \Gamma_r(x+1) =\lim_{r \to \infty} \dfrac{r! r^{x+1}}{(x+1)...(x+r)(x+r+1)}$

$\Gamma(x+1) = \lim_{r \to \infty} \dfrac{x r! r^{x}r}{x(x+1)...(x+r)(x+r+1)}$

$\Gamma(x+1) = \lim_{r \to \infty} \dfrac{r! r^x}{x(x+1)...(x+r)} x\dfrac{r}{x+r+1}$

$\Gamma(x+1) = x \lim_{r \to \infty} \dfrac{r!r^x}{x(x+1)...(x+r)} \lim_{ r \to \infty} \dfrac{x}{x+r+1}$

$\Gamma(x+1) = x \Gamma(x)$

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