I have made the following proof for convergence of gamma function.
Please tell me if it is correct.
$\int_0^\infty e^{-x}x^{n-1}dx$ converges for all $n>0.$
Step I: $\int_0^1 e^{-x}x^{n-1}dx$ exists $\forall~n>0:$ Clearly $\int_0^1 e^{-x}x^{n-1}dx$ is Riemann integrable for $n\ge1.$
For $0<n<1,$ $0$ is the only point of infinite discontinuity of $e^{-x}x^{n-1}$ on $[0,1].$
Since $0<e^{-x}x^{n-1}\le x^{n-1}$ on $(0,1]$ and $\int_0^1 x^{n-1}dx$ is convergent for $0<n<1,$ $\int_0^1 x^{n-1}dx$ exists for $0<n<1.$
Step II: $\int_1^\infty e^{-x}x^{n-1}dx$ exists $\forall~n>0:$ Choose $N>1$ such that $e^{x/2}>x^{n-1}$ for $x>N.$ Thus $e^{-x}x^{n-1}<e^{-x/2}$ for $x>N.$
Since $\int_N^\infty e^{-x/2}dx$ is convergent, so is $\int_N^\infty e^{-x}x^{n-1}dx.$
On the other hand, $\int_1^N e^{-x}x^{n-1}dx$ clearly exists. Thus $\int_1^\infty e^{-x}x^{n-1}dx$ exists.
From Step I and II, it is clear that $\int_0^\infty e^{-x}x^{n-1}dx$ converges for all $n>0.$