Letting n be a natural number and applying integration by parts one time gives
\begin{align*}
\Gamma(n+1) & = \int_0^{\infty} u^n e^{-u} du\\
& = -u^n \cdot e^{-u} \big |_0^{\infty} + n \int_0^{\infty} u^{n-1} e^{-u} du \\
& = 0 - 0 + n \Gamma(n)
\end{align*}
Continuing using an inductive argument to obtain the final result.