Introduction : In this solution, we write formula of Gamma function and prove that prove that Gamma(n+1)=nGamma(n) in the following
Question : Prove that
\[\mathrm{\Gamma}\left(n+1\right)=n\mathrm{\Gamma}\left(n\right)\]
Solution :
To prove
\[\mathrm{\Gamma}\left(n+1\right)=n\mathrm{\Gamma}\left(n\right)\]
We general formula of Gamma function
\[\mathrm{\Gamma}\left(n\right)=\int_{0}^{\infty}{e^{-x}x^{n-1}dx}\]
replacing n by n+1 here we get
\[\mathrm{\Gamma}\left(n+1\right)=\int_{0}^{\infty}{e^{-x}x^{n+1-1}dx}\]
which becomes
\[=\int_{0}^{\infty}{e^{-x}x^ndx}\]
now integrating by parts
\[={\left[-e^{-x}x^n\right]^\infty}_0-\int_{0}^{\infty}\left\{\frac{d}{dx}\left(x^n\right)\int{e^{-x}dx}\right\}dx\]
\[=-\left(limit_{x\rightarrow\infty}\frac{x^n}{e^x}-0\right)+n\int_{0}^{\infty}{{x}^{n-1}e^{-x}dx}\]
\[=-\left(0-0\right)+n\int_{0}^{\infty}{e^{-x}x^{n-1}dx}\]
\[\mathrm{\Gamma}\left(n+1\right)=n\int_{0}^{\infty}{e^{-x}x^{n-1}dx}\]
using Gamma of n in the right hand side
\[=n\mathrm{\Gamma}\left(n\right)\]
This is proved that
\[\mathrm{\Gamma}\left(n+1\right)=n\mathrm{\Gamma}\left(n\right)\]