Fix $z \in Z$. Applying integration by parts, one has \begin{equation} \begin{split} \Gamma(z + 1) & = \int^{\infty}_0 t^z e^{- t} \, d t \\ & = \left[ - t^z e^{- t} \right]^{\infty}_0 + z \int^{\infty}_0 t^{z - 1} e^{- t} \, d t \\ & = \lim_{b \to \infty} (- b^z e^{- b}) + 0 + z \Gamma(z) \\ & = z \Gamma(z) \end{split} \end{equation} $\square$