I assume that $n$ is a real number. Split the gamma improper integral $\Gamma(n)=\int_{0}^{\infty}e^{-x}x^{n-1}dx\tag{0}$ into $I_1+I_2$, where $I_1=\int_{0}^{1}e^{-x}x^{n-1}dx\tag{1}$ and $I_2=\int_{1}^{\infty}e^{-x}x^{n-1}dx\tag{2}$
- To prove that the integral $I_2$ is always convergent use the fact that for any real number $\alpha $ the integral $ \int_{1}^{\infty }e^{-x}x^{\alpha }dx\tag{3}$ is convergent, by the limit comparison test $\lim_{x\rightarrow \infty }\frac{e^{-x}x^{\alpha }}{x^{-2}}=0\tag{4}$ with the convergent integral $\displaystyle\int_{1}^{\infty }\dfrac{dx}{x^{2}}\tag{5}.$
- As for $I_1$ consider two cases. (a) If $n\geq 1$ observe that $\lim_{x\rightarrow 0}e^{-x}x^{n-1}=0$, so $I_1$ is a proper integral. (b) If $0, the integrand $e^{-x}x^{n-1}$ behaves like $x^{n-1}$ near $n=0$, because $e^{-x}\rightarrow 1$ as $x\rightarrow 0$. Since $\displaystyle\int_{0}^{1}\dfrac{dx}{x^{1-n}}\tag{6}$ is convergent if and only if $1-n<1$, i.e. $n>0$, so is $I_1$.
It follows that $\Gamma(n)=I_1+I_2$ is convergent for $n>0.$