Let $V_m(r)=\operatorname{vol}B(0,r)$ and prove $V_{n+1}(r) = 2V_n(1)\int_0^r(r^2-t^2)^{n/2}\,\mathrm dt$ 
Let $V_{m}(r)$ the volume of the ball with center in origin and radius $r$ in $\mathbb{R}^{m}$. Prove the inductive relation $V_{n+1}(r) = 2V_{n}(1)\int_{0}^{r}(r^{2}-t^{2})^{n/2}\,\mathrm dt$ and conclude that
  $$V_{m}(r) = \frac{r^{m}\pi^{m/2}}{\left(\frac{m}{2}\right)!}$$
  when $m$ is even and
  $$V_{m}(r) = r^{m}\pi^{(m-1)/2}\frac{2^{m}\left(\frac{m-1}{2}\right)!}{m!}$$
  if $m$ is odd. Note that of these formulas results a curious fact:
  $$\lim_{m \to \infty}V_{m}(r) = 0.$$

For $V_{n+1}(r) = 2V_{n}(1)\int_{0}^{r}(r^{2}-t^{2})^{n/2}\,\mathrm dt$, I checked the case $n=1$. For general case, I think I should apply a change of variables, but I could not do it correctly. Can anybody help me? Please, only with this equality because I want to try to do the other parts on my own.

Edit. I try to use the Oliver hint, but I could not complete the proof.
 A: $\def\peq{\mathrel{\phantom{=}}{}}\def\d{\mathrm{d}}$For any $n \geqslant 1$ and $(x_1, \cdots, x_{n + 1}) \in \mathbb{R}^{n + 1}$, note that\begin{align*}
&\mathrel{\phantom{\Longleftrightarrow}} (x_1, \cdots, x_{n + 1}) \in B_n(0, r)\\
&\Longleftrightarrow \sum_{k = 1}^{n + 1} x_k^2 < r^2 \Longleftrightarrow -r < x_{n + 1} < r \text{ and } \sum_{k = 1}^n x_k^2 < r^2 - x_{n + 1}^2\\
&\Longleftrightarrow -r < x_{n + 1} < r \text{ and } (x_1, \cdots, x_n) \in B_n(0, \sqrt{\smash[b]{r^2 - x_{n + 1}^2}})
\end{align*}
and $V_n(r) = r^n V_n(1)$, thus\begin{align*}
V_{n + 1}(r) &=  \int\limits_{B_{n + 1}(0, r)} \d x = \int_{-r}^r \d x_{n + 1} \int\limits_{B_n(0, \sqrt{\smash[b]{r^2 - x_{n + 1}^2}})} \d x\\
&= \int_{-r}^r V_n(\sqrt{\smash[b]{r^2 - x_{n + 1}^2}}) \,\d x_{n + 1} = \int_{-r}^r (r^2 - x_{n + 1}^2)^{\frac{n}{2}} V_n(1) \,\d x_{n + 1}\\
&= V_n(1) \int_{-r}^r (r^2 - x_{n + 1}^2)^{\frac{n}{2}} \,\d x_{n + 1} = 2V_n(1) \int_0^r (r^2 - t^2)^{\frac{n}{2}} \,\d t.
\end{align*}
Now for any $n \geqslant 1$,$$
I_n := \int_0^1 (1 - t^2)^{\frac{n}{2}} \,\d t = \int_0^{\tfrac{π}{2}} (1 - \sin^2 θ)^{\frac{n}{2}} \,\d(\sin θ) = \int_0^{\tfrac{π}{2}} \cos^{n + 1} θ \,\d θ,
$$
and $I_1 = \dfrac{π}{4}$, $I_2 = \dfrac{2}{3}$. For $n \geqslant 3$, because\begin{align*}
I_n &= \int_0^{\tfrac{π}{2}} \cos^{n + 1} θ \,\d θ = \int_0^{\tfrac{π}{2}} \cos^n θ \,\d(\sin θ)\\
&= \cos^n θ \sin θ \biggr|_0^{\tfrac{π}{2}} - \int_0^{\tfrac{π}{2}} \sin θ \,\d(\cos^n θ)\\
&= n \int_0^{\tfrac{π}{2}} \sin^2 θ\cos^{n - 1} θ \,\d θ = n \int_0^{\tfrac{π}{2}} (1 - \cos^2 θ) \cos^{n - 1} θ \,\d θ\\
&= n(I_{n - 2} - I_n),
\end{align*}
then $I_n = \dfrac{n}{n + 1} I_{n - 2}$. Thus for $k \geqslant 1$,\begin{gather*}
I_{2k - 1} = I_1 \prod_{j = 2}^k \frac{2j - 1}{2j} = \frac{π}{2} \cdot \frac{(2k - 1)!!}{(2k)!!},\quad I_{2k} = I_2 \prod_{j = 2}^k \frac{2j}{2j + 1} = \frac{(2k)!!}{(2k + 1)!!}.
\end{gather*}
Therefore for $k \geqslant 1$,\begin{align*}
V_{2k - 1}(1) &= 2V_{2k - 2}(1) I_{2k - 2} = \cdots = 2^{2k - 2} V_1(1) \prod_{j = 1}^{2k - 2} I_j\\
&= 2^{2k - 1} \cdot \left( \frac{π}{2} \right)^{k - 1} \frac{1}{(2k - 1)!!} = π^k \cdot \frac{2^{2k - 1} (k - 1)!}{(2k - 1)!},
\end{align*}\begin{align*}
V_{2k}(1) &= 2V_{2k - 1}(1) I_{2k - 1} = \cdots = 2^{2k - 1} V_1(1) \prod_{j = 1}^{2k - 1} I_j\\
&= 2^{2k} \cdot \left( \frac{π}{2} \right)^k \frac{1}{(2k)!!} = \frac{π^k}{k!},
\end{align*}
and$$
V_{2k - 1}(r) = r^{2k - 1} V_{2k - 1}(1) = π^k r^{2k - 1} \cdot \frac{2^{2k - 1} (k - 1)!}{(2k - 1)!},\\
V_{2k}(r) = r^{2k} V_{2k}(1) = \frac{π^k r^{2k}}{k!}.
$$
A: The idea is to use Fubini, i.e., slice integration. Let $B_n(r)$ denote the open unit ball in $\mathbb R^n$ with center at the origin and radius $r.$ Let $m_n$ the the Lebesgue volume measure on $\mathbb R^n.$ Recall that $m_n(B_n(r)) = r^nm_n(B_n(1)).$
We can view $B_{n+1}(r)$ as the set $\{(t,x): t\in (-r,r), x\in \mathbb R^n,|x|<(r^2-t^2)^{1/2}\}.$ With this setup, Fubini gives
$$m_{n+1}(B_{n+1}(r)) = \int_{-r}^r m_n(B_n((r^2-t^2)^{1/2}))\,dt$$ $$ = \int_{-r}^r [(r^2-t^2)^{1/2}]^nm_n(B_n(1))\,dt = m_n(B_n(1))\int_{-r}^r (r^2-t^2)^{n/2}\,dt.$$
The last expression is precisely what we want to see, because $V_n(1)=m_n(B_n(1)),$ and the integrand is even, which implies $\int_{-r}^r = 2\int_{0}^r.$
You said you checked the result when $n=1.$ But to get comfortable with this technique, you might try it for the case $n=2$ at first, where again you know the answer.
A: We have that
\begin{align}
V_{n+1}(r) &= \int_{-r}^r V_n(r(t)) dt \hskip{1cm} (\text{where}\hskip{0.5cm} r(t)=\sqrt{r^2-t^2}\> ) \\
           &= 2\int_0^r V_n(1) [r(t)]^n dt \hskip{1cm} (\text{using} \hskip{0.5cm} V_n(r)=V_n(1)r^n \>) \\
           &= 2V_n(1)\int_0^r  (r^2-t^2)^{n/2} dt \\
\end{align}
