Volume of n dimensional ball 
*

*The open ball of radius $r$ in  $\mathbb{R}^{N}$ is the set
$\left\{\left(x_{1}, x_{2},\ldots,x_{n}\right)\in \mathbb{R}^{N} \mid \sum_{i = 1}^{N} x_{i}^{2} < r^{2}\right\}$.

*By definition its volume $V_N\left(r\right) =
\int\int\cdots\int 1\,\,{\rm d}x_{1}\,{\rm d}x_{2}\ldots
{\rm d}x_{N}$
$$
\mbox{How to prove that}\quad 
V_{N} = V_{N - 1}\int_{0}^{\pi/2}\cos^{n}\left(\theta\right)
\,{\rm d}\theta\,\,\, ?
$$
$V_{n}$ is the volume of the $n$-dimensional unit ball.
Any idea how I can show this please. I have no idea what sort of approach I should take.
 A: Search:
"Finding Volume and Surface Area of Hyperspheres in ${\Bbb R}^n$" (Mario Sracic)
"The volume of a n-dimensional hypersphere" (A. E. Lawrence)
"The volume of a n-dimensional sphere in ${\Bbb R}^{n+1}$"
A: $$
V_n(r) = \int_0^r \int_{x_1^2+\cdots+x_{n-1}^2\le r^2-x^2} dx_1\ldots dx_{n-1} dx
\\ = \int_0^r  V_{n-1}(\sqrt{r^2-x^2}) dx\\
= \int_0^{\frac\pi 2}  V_{n-1}(r\cos\theta) r\cos\theta \ \ d\theta
$$
Now use the fact that $V_n(r) = r^n V_n(1) =: r^n V_n$:
$$
r^n V_n 
= \int_0^{\frac\pi 2}  V_{n-1}(r\cos\theta)^{n-1} r\cos\theta \ \ d\theta\\
V_n 
= V_{n-1}\int_0^{\frac\pi 2}  (\cos\theta)^{n}\ d\theta
$$
A: As a matter of fact, you cannot prove that
$$ V_n \stackrel?= V_{n-1} \int _0 ^{\pi/2} \cos^n \theta \, \mathrm d\theta.$$
The two sides of this alleged "equation" are not equal.
For example, let $n = 3.$ Then $V_{n-1} = V_2 = \pi$
and
$$
\int_0^{\pi/2} \cos^n \theta \, \mathrm d\theta
= \int_0^{\pi/2} \cos^3 \theta \, \mathrm d\theta = \frac23.
$$
Hence
$$
V_{n-1} \int _0 ^{\pi/2} \cos^n \theta \, \mathrm d\theta = \frac23\pi.
$$
But
$$
V_n = V_3 = \frac43\pi.
$$
To correct the equation, we can either integrate from $-\frac\pi2$ to $\frac\pi2$ instead of $0$ to $\frac\pi2$
(equivalent to integrating in Cartesian coordinates from $-r$ to $r$ instead of from $0$ to $r$, as suggested in a comment under
another answer)
or we can recognize the symmetry of that integral and simply double the original integral. That is,
$$
V_n = V_{n-1} \int _{-\pi/2} ^{\pi/2} \cos^n \theta \, \mathrm d\theta
=  2 V_{n-1} \int _0 ^{\pi/2} \cos^n \theta \, \mathrm d\theta.
$$

A correct derivation in the style of
the answer mentioned above
could be
\begin{align}
V_n
&= \int_{x_1^2+\cdots+x_{n-1}^2+x_n^2\leq 1}
      \mathrm dx_1 \cdots \mathrm dx_{n-1}\, \mathrm dx_n \\
&= \int_{x_n^2\leq 1} \int_{x_1^2+\cdots+x_{n-1}^2\le 1-x^2}
      \mathrm dx_1\cdots \mathrm dx_{n-1}\, \mathrm dx_n \\
&= \int_{-1}^1 \left(\sqrt{1-x^2}\right)^{n-1} V_{n-1}\, \mathrm dx \\
&= \int_{-\frac\pi2}^{\frac\pi2} (\cos^{n-1}\theta) V_{n-1}\cdot \cos\theta
      \,\mathrm d\theta \\
&= V_{n-1} \int_{-\frac\pi2}^{\frac\pi2} \cos^n\theta \,\mathrm d\theta
\end{align}
using the notational convention that the volume of an $n$-ball of radius $r$ is $r^n V_n.$
