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
 A: 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 $\Gamma(0 + 1) = 0!$.
A: You can make it short:
From the definition of $\Gamma$, we have
$$\Gamma(1)=\int_0^\infty e^{-x}\,dx=1.$$
Then by parts, integrating $e^{-x}$,
$$\Gamma(n+1)=\int_0^\infty x^{n}e^{-x}\,dx=-\left.x^ne^{-x}\right|_0^\infty+n\int_0^\infty x^{n-1}e^{-x}\,dx=n\Gamma(n)$$
because $\lim_{x\to\infty}x^ne^{-x}=0$. The integral converges for all positive integer $n$.
This shows that $\Gamma(n+1)$ and $n!$ follow the same recurrence and are equal for all $n$.

The crux of the proof is the integration by parts, which reduces the exponent of $x$ and induces the recurrence relation.

A direct proof:
From $(P(x)e^{-x})'=(P'(x)-P(x))e^{-x}$, you see that you can find the antiderivative of $x^ne^{-x}$ by solving
$$P'(x)-P(x)=x^n.$$
Expanding the polynomial, this is
$$-p_nx^n+(np_n-p_{n-1})x^{n-1}+((n-1)p_{n-1}-p_{n-2})x^{n-2}+\cdots p_1-p_0=x^n$$
and by identification,
$$p_k=-\frac{n!}{k!}.$$
Hence
$$\int_0^\infty x^ne^{-x}dx=-n!\left.\left(\frac{x^n}{n!}+\frac{x^{n-1}}{(n-1)!}+\cdots1\right)e^{-x}\right|_0^\infty=n!$$ as only the last term contributes.
A: 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 
