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 $00$, so is $I_1$.
It follows that $\Gamma(n)=I_1+I_2$ is convergent for $n>0.$