For $t>0$ I want to show that $$\frac{1}{(4\pi t)^{n/2}}\int_{\mathbb{R}^{n}}e^{\frac{-\|x\|^{2}}{4t}}dx=1$$
So far, I have
$$\begin{aligned} \frac{1}{(4\pi t)^{n/2}}\int_{\mathbb{R}^{n}}e^{\frac{-\|x\|^{2}}{4t}}dx&=\frac{1}{(4\pi t)^{n/2}}\int_{\mathbb{S}^{n-1}}\int_{0}^{\infty}r^{n-1}e^{-\frac{r^{2}}{4t}}drd\theta \\ &=\frac{\omega_{n-1}}{(4\pi t)^{n/2}}\int_{0}^{\infty}r^{n-1}e^{-\frac{r^{2}}{4t}}dr \\ &=\frac{2\pi^{n/2}}{(4\pi t)^{n/2}\Gamma\left(\frac{n}{2}\right)}\int_{0}^{\infty}r^{n-1}e^{-\frac{r^{2}}{4t}}dr \end{aligned}$$
But from here I have not been able to proceed, although I realise that the integral looks something like the integral function.
Although the Wikipedia article on Gaussian integrals shows computations on $\mathbb{R}$ and $\mathbb{R}^{2}$ I could not find resources on $\mathbb{R}^{n}$.
Get rid of $t$ through a suitable substitution then apply Fubini's theorem, without switching to polar coordinates: $$ \iiint_{\mathbb{R}^3} e^{-(x^2+y^2+z^2)}\,dx\,\,dy\,dz = \left(\int_{-\infty}^{+\infty}e^{-u^2}\,du\right)^3,$$ for instance.