Volume of a hypersphere

There is a standard "trick" for evaluating this. First, note that $$ \int_{\mathbb{R}}e^{-\pi x^2}\,\mathrm{d}x=1\tag{1} $$ Multiplying $(1)$ together $n$ times, we get $$ \int_{\mathbb{R}^n}e^{-\pi x^2}\,\mathrm{d}x=1\tag{2} $$ Converting $(2)$ to polar coordinates yields $$ \int_0^\infty\omega_{n-1}e^{-\pi r^2}r^{n-1}\,\mathrm{d}r=1\tag{3} $$ where $\omega_{n-1}$ is the area of the $n-1$ dimensional unit sphere. Compute $\omega_{n-1}$ as follows $$ \begin{align} 1 &=\int_0^\infty\omega_{n-1}e^{-\pi r^2}r^{n-1}\,\mathrm{d}r\\ &=\frac12\int_0^\infty\omega_{n-1}e^{-\pi r^2}r^{n-2}\,\mathrm{d}r^2\\ &=\frac{\pi^{-n/2}}2\int_0^\infty\omega_{n-1}e^{-s}s^{n/2-1}\,\mathrm{d}s\\ &=\frac{\pi^{-n/2}}2\omega_{n-1}\Gamma(n/2)\\ \omega_{n-1}&=\frac{2\pi^{n/2}}{\Gamma(n/2)}\tag{4} \end{align} $$ Now, we can compute the volume of the $n$ dimensional sphere, by $$ \begin{align} \int_0^r\omega_{n-1}t^{n-1}\,\mathrm{d}t &=\frac{\omega_{n-1}}{n}r^n\\ &=\frac{2\pi^{n/2}}{n\Gamma(n/2)}r^n\\[6pt] &=\left\{\begin{array}{} \frac{\pi^{n/2}}{(n/2)!}r^n&\quad\text{if $n$ is even}\\[6pt] \frac{2^n\left(\frac{n-1}{2}\right)!}{n!}\pi^{\frac{n-1}{2}}r^n&\quad\text{if $n$ is odd} \end{array}\right.\tag{5} \end{align} $$