$0! `:=` \Gamma (1)$
$(z) := _{0}^{} t{z-1}e{-t}dt $
Γ(1) = ∫0∞t0e−tdt
= ∫0∞e−tdt
using the substitution method for integration.
$x `:=` -t$
$\implies \frac{dx}{d t} = -1$
⟹ dx = −dt
And the bounds of the integral are now 0 to −∞.
⟹ ∫0∞e−tdt = −∫0−∞exdx
We know that the integral of ex is itself, ex.
= −[ex]0−∞
$ = - (0 - 1) = - (-1) = 1$. QED.