The Gamma function interpolates the factorial

Step 1 (Base case). $\Gamma(1) = \int_0^\infty t^0 e^{-t}\,dt = \int_0^\infty e^{-t}\,dt = \big[-e^{-t}\big]_0^\infty = 0-(-1) = 1 = 0!$, so the formula holds for $n=0$.

Step 2 (Recurrence by integration by parts). For $\operatorname{Re}(z) > 0$, integrate $\Gamma(z)=\displaystyle\int_0^\infty t^{z-1}e^{-t}\,dt$ by parts with $u=t^z$, $dv=e^{-t}\,dt$: $\Gamma(z+1) = \int_0^\infty t^z e^{-t}\,dt = \big[-t^z e^{-t}\big]_0^\infty + z\int_0^\infty t^{z-1}e^{-t}\,dt = 0 + z\,\Gamma(z)$, using that the boundary term vanishes at both ends. This proves $\Gamma(z+1) = z\,\Gamma(z)$.

Step 3 (Induction). Assume $\Gamma(k+1) = k!$ for some non-negative integer $k$. By Step 2 with $z=k+1$, $\Gamma(k+2) = (k+1)\,\Gamma(k+1) = (k+1)\cdot k! = (k+1)!$.

Step 4 (Conclude). Since the base case $n=0$ holds (Step 1) and the inductive step carries the identity from $n=k$ to $n=k+1$ (Step 3), $\Gamma(n+1)=n!$ holds for every non-negative integer $n$ by mathematical induction.