Show that $\int_{0}^{+\infty}e^{-\sqrt[n]{v}}\text{d}v=n!$.

Question

I want to know how to prove that $$ \int_{0}^{+\infty}e^{-\sqrt[n]{v}}\text{d}v=n! $$ I wanted to use that $$ \int_{0}^{+\infty}t^{n}e^{-t}\text{d}t=n! $$ and use $v=t^n$ but it didn't work because I obtained that it is equal to $$ \frac{1}{n}\int_{0}^{+\infty}v^{1/n}e^{-\sqrt[n]{v}}\text{d}v $$

Answer 1

It works well

$v=t^n$ then $ dv=nt^{n-1}dt$

And $ e^{-\sqrt[n]{v}}=e^{-t}$

$$I=n\int_0^{\infty}e^{-t}t^{n-1}dt=n\Gamma(n)=n!$$

Answer 2

Your proposed substitution is $t=\sqrt[n]{v}=v^{\frac1n}$. Then $v=t^n$, $t=0$ when $v=0$, $t=+\infty$ when $v=+\infty$ and

$$dt=\frac1n v^{1-\frac1n}\,dv=\frac1n\,\frac1{\sqrt[n]v}v\,dv,$$

that is, $dt=\frac1n\,\frac1t\,t^n\,dv\implies dv=nt^{n-1}\,dt$. So, after the substitution, you get

$$ \int_{0}^{+\infty}e^{-\sqrt[n]{v}}\text{d}v=\int_0^{+\infty}e^{-t}\,nt^{n-1}\,dt $$

Can you take it from here?