Calculate the volume of the $n$-dimensional ball using change of variables.
My attempt.
Lemma 1. For each $k \in \mathbb{N}$ we have $$\int_{0}^{\pi}\sin^{k}xdx = \frac{k-1}{k}\int_{0}^{\pi}\sin^{k-2}xdx.$$
Proof. Done.
Corollary. If $k$ is even, then $$\int_{0}^{\pi}\sin^{k}xdx = \left(\frac{k-1}{k}\right)\left(\frac{k-3}{k-2}\right)\cdots\frac{3}{4}\int_{0}^{\pi}\sin^2xdx,$$ $$\int_{0}^{\pi}\sin^{k-1}xdx = \left(\frac{k-2}{k-1}\right)\left(\frac{k-4}{k-3}\right)\cdots\frac{2}{3}\int_{0}^{\pi}\sin xdx.$$ Moreover, $$\int_{0}^{\pi}\sin^{k}xdx\int_{0}^{\pi}\sin^{k-1}xdx = \frac{2\pi}{k}.$$
Proof. Done.
Thus, if $k$ is even, take $I_{k} = \int_{0}^{\pi}\sin^{k}xdx$, we have $$I_{k}I_{k-1}\cdots I_{1} = (I_{k}I_{k-1})\cdots (I_{2}I_{1}) = \frac{(2\pi)^{\frac{k}{2}}}{k(k-2)\cdots 2}.$$ Analogously, if $k$ is odd, $$I_{k}I_{k-1}\cdots I_{1} = (I_{k}I_{k-1})\cdots (I_{3}I_{2})I_{1} = \frac{2(2\pi)^{\frac{k-1}{2}}}{k(k-2)\cdots 3}.$$
Now, $$\begin{eqnarray*}\mathrm{vol}(B_{R}) & = & \int_{B_{R}}dx\\ & = & \int_{(0,R)\times(0,\pi)^{n-2}\times(0,\pi)}r^{n-1}\sin_{n-2}(\theta_{1})\cdots\sin(\theta_{n-1})drd\theta_{1}...d\theta_{n-1}d\theta_{n}\\ & \underbrace{=}_{\color{red}{\mathrm{Fubini}}} & \int_{0}^{R}r^{n-1}dr\int_{0}^{2\pi}d\theta_{n}\int_{0}^{\pi}\sin^{n-2}(\theta_{1})d\theta_{n-1}\cdots\int_{0}^{\pi}\sin(\theta_{n-1})d\theta_{n-1}\\ & = & 2\pi\frac{R^{n}}{n}I_{n-2}I_{n-3}\cdots I_{1}. \end{eqnarray*}$$
By previous comments, $$I_{n-2}I_{n-3}\cdots I_{1} = \frac{(2\pi)^{\frac{n-2}{2}}}{(n-2)\cdots 2}$$ if $n$ is even, and $$I_{n-2}I_{n-3}\cdots I_{1} = \frac{2(2\pi)^{\frac{n-3}{2}}}{(n-2)\cdots 3}$$ if $n$ is odd.
Therefore, $$\mathrm{vol}(B_{R}) = \begin{cases} 2\pi\frac{R^{n}}{n}\frac{(2\pi)^{\frac{n-2}{2}}}{(n-2)\cdots 2}& \text{if }n\text{ is even} \\ 2\pi\frac{R^{n}}{n}\frac{2(2\pi)^{\frac{n-3}{2}}}{(n-2)\cdots 3}& \text{if }n\text{ is odd}\end{cases},$$ that is, $$\mathrm{vol}(B_{R}) = \begin{cases} \frac{(2\pi)^{\frac{n}{2}}R^{n}}{n(n-2)\cdots 2}& \text{if }n\text{ is even} \\ \frac{2(2\pi)^{\frac{n-1}{2}}R^{n}}{n(n-2)\cdots 3}& \text{if }n\text{ is odd}\end{cases}.$$
Is this correct? Testing for $n = 3$, is right. I think that the final result is right, but I'm not sure about all the details in the last part.