We know that $$\Gamma(n) \equiv \int_0^{\infty}t^{n-1}e^{-t} \textrm{d}t = (n-1)!$$ But this just looks like another formula and I can't see why this would be equal to $(n-1)!$.
Is there a proof that $$\Gamma(n) = (n-1)!$$ ?
I'm not too familiar with the Gamma function. Any help is appreciated.
On
Just to give another (non-rigorous) approach, if $|x|\lt1$, then, slipping an infinite sum inside the integral, we have
$$\sum_{n=0}^\infty{\Gamma(n+1)\over n!}x^n=\int_0^\infty\left(\sum_0^\infty{(tx)^n\over n!}\right)e^{-t}dt=\int_0^\infty e^{-t(1-x)}dt={1\over1-x}=\sum_{n=0}^\infty x^n$$
and so $\Gamma(n+1)/n!=1$ for all $n$.
To make this rigorous would require justifying two key steps: the interchange $\sum\int=\int\sum$, and the conclusion that $\sum a_nx^n=\sum b_nx^n$ for all $|x|\lt1$ implies $a_n=b_n$ for all $n$.
Let $\epsilon>0$ then integrating by parts for all $n \in \mathbb{N}^{*}$ $$ \int_{0}^{\epsilon}t^{n}e^{-t}\text{d}t=\left[-e^{-t}t^{n}\right]^{\epsilon}_{0}+\int_{0}^{\epsilon}nt^{n-1}e^{-t}\text{d}t $$ And $\displaystyle -e^{-\epsilon}\epsilon^{n} \underset{\epsilon \rightarrow +\infty}{\rightarrow}0$ so when $\epsilon \rightarrow +\infty$ it becomes $$ \Gamma\left(n+1\right)=n\Gamma\left(n\right) $$ And $$ \Gamma\left(1\right)=\int_{0}^{+\infty}e^{-t}\text{d}t=1 $$ With $0!=1$ and $n!=n\left(n-1\right)!$ both expressions satisfies exactly the same induction relations. What is more interesting is that $\Gamma$ satisfies it for all $x \in \mathbb{R}^{*+}$.