TITLE: Proof that the gamma function is an extension of the factorial function
QUESTION [5 upvotes]: I've already proved that $$\Gamma (n)= (n-1)!$$  but I don´t really know what else to do to verify that $\Gamma$ is an extension of the factorial function for real numbers (positive) Thank you! And I´m sorry for my language, I am Spanish, so thank you again for trying to understand me.

REPLY [4 votes]: Consider $$\Gamma (z)=\int_0^\infty e^{-t}t^{z}{dt\over t}.$$ We can rewrite this as $$\Gamma (z)=\int_0^1 e^{-t}t^{z}{dt\over t}+\int_1^\infty e^{-t}t^{z}{dt\over t}.$$ In the first term of this sum we see that the power series representation of $e^{-t}$ converges uniformly which implies that the series can be integrated term by term. So, $$\Gamma (z)=\int_0^1\sum_{n=0}^\infty {(-1)^n\over n!}t^{z+n}{dt\over t}+\int_1^\infty e^{-t}t^{z}{dt\over t}=\sum_{n=0}^\infty {(-1)^n\over n!(z+n)}+\int_1^\infty e^{-t}t^{z}{dt\over t}.$$ We can see that the series converges for $z\neq 0, -1, -2,...$ which is a meromorphic function. Its poles are simple poles at the non-positive integers. The residue at $-n$ is $(-1)^n\over n!$. The last integral extends as an entire function of z.
Thus $\Gamma (z)$ has been analytically continued to the entire complex plane except for $z\neq0,1,2,...$.