The Gamma function and factorial satisfy $\Gamma(n+1) = n!$ This was a question from a mock paper, for my upcoming exam however my teacher unhelpfully did not post any solutions.


  Prove that $\Gamma(n+1)=n!$.




Can anyone check if my proof is correct.

Thank you for reading.

-Alexis
Note that he property $$G(n + 1) = n G(n)$$ you establish also holds for any constant multiple of $\Gamma$, including the zero function.

Since the proof you give is basically an inductive argument (it might be useful to say a little more in your solution about how this goes), it suffices to add a base case, that is, show that the identity holds for the lowest applicable value of $n$. Since $0!$ makes sense (but $(-1)!$ is not defined), one should show that $\Gamma(0 + 1) = 0!$.
If  $~n~$ is a positive integer$$\Gamma(n+1)=n!$$ For negative integers it's undefined.

So that.                              $$\Gamma(n+1)= n \Gamma(n)$$
            $$=n \Gamma(n-1+1)=n (n-1)\Gamma(n-1) $$
           $$ =n(n-1)\Gamma(n-2+1)   $$
          $$  =n(n-1)(n-2)\Gamma(n-2)   $$
         $$  =.......   ....   ..... $$
         $$   =n(n-1)(n-2)....3.2.1.\Gamma(1)$$
         $$   =n(n-1)(n-2)..... 3.2.1. $$
        $$    = n!   $$ since $\Gamma(1)=1$,  

It's simple proof