Uni-numeral calculus

The integral $\int^\infty_0 \frac{t^n}{e^t}\,dt$

can be written as $\int^\infty_0 t^{n+1-1}e^{-t}\,dt= \Gamma(n+1)$

& as n is an positive integer $\Gamma(n+1) = (n)!$

I will prove this theorem by induction.

Base case ( $n=0$ ): $\int_0^\infty t^0e^{-t}dt=-e^{-t}\Big|_0^\infty=-(0-1)=1=0!$

Induction step: Assume $\int_0^\infty t^ne^{-t}dt=n!.$

I now show $\int_0^\infty t^{n+1}e^{-t}dt=(n+1)!.$

$\int_0^\infty t^{n+1}e^{-t}dt=-t^{n+1}e^{-t}\Big|_0^\infty+\int_0^\infty (n+1)t^ne^{-t}dt=0+(n+1)n!=(n+1)!.$

By the principle of mathematical induction, $\int_0^\infty t^ne^{-t}dt=n!$ for all $n\geq 0.$