Show $\int_\mathbb{R^n} e^{-||x||^2}d^n x = \pi^{n/2}$

Question

Let $||x||:= \sqrt{x_1^2+...+x_n^2}$ be the euclidean norm of a vector $x\in\mathbb{R^n}$. Show that $$\int_\mathbb{R^n} e^{-||x||^2}d^n x = \pi^{n/2}$$

Answer 1

Hint: $$e^{-\|x\|^2} = e^{-x_1^2} \ldots e^{-x_n^2}$$

Answer 2

By Fubini's Theorem, \begin{align*} \int_{{\bf{R}}^{n}}e^{-|x|^{2}}dx=\left(\int_{{\bf{R}}}e^{-x^{2}}dx\right)^{n}. \end{align*}