$\Omega_n : = \{ (x_1, \cdots , x_n) \in \mathbb{R}^n \ | \ |x_1|+ \cdots +|x_n| \leqq r \}$
$V_n$ : Volume of $\Omega_n$ $\left( \text{therefore}, V_n = \int_{\Omega} 1 \ d(x_1, \cdots , x_n) \right) $
Problem : \begin{equation} \text{Prove that }V_n \left(= \int_{\Omega} 1 \ d(x_1, \cdots , x_n) \right) = \dfrac{(2r)^n}{n \ !} \cdots \ast_1. \end{equation}
(I calculated $V_1$ and $V_2$, and I derived $V_1=2r, V_2=2r^2$. These machies with $\ast_1$.)
Probably, I might be able to prove $\ast_1$ with mathematical induction., but I don't know how I have to work.
I have to suppose $V_{n-1}=\dfrac{(2r)^{n-1}}{(n-1)\ !}$ holds, and calculate $V_n$.
I don't know how I have to use $V_{n-1}=\dfrac{(2r)^{n-1}}{(n-1)\ !}$
in calculating $V_n$.
I would like you to tell me some ideas.
For the purposes of this proof, it might help to add the number $r$ to the notation (in case you want to use different values of $r$ at different times). I suggest writing $$ \Omega_n(r) = \left\{(x_1,\dots,x_n) ~\middle|~ \sum_{i=1}^n |x_i| = r \right\} $$
I assume you are familiar with the fact that you can compute volume integrals "by slicing". Meaning if you take the cross section of $\Omega$ in the hyperplane $x_n = a$, and if that cross section has $(n-1)$-dimensional volume $v(a)$, then $$ V_n(r) = \int_{-r}^r v(a) \, da $$ In this case, the cross-sectional volume is $v(a) = V_{n-1}(r-a)$, since the cross-sectional slice just looks like the set $\Omega_{n-1}(r-a)$. Putting this all together, you have $$ \begin {align*} \int_{\Omega_n(r)} dx_1 \cdots dx_n &= \int_{-r}^r \left( \int_{\Omega_{n-1}(r-x_n)} dx_1 \cdots dx_{n-1} \right) \, dx_n \end {align*} $$
Now you can use induction to evaluate the inner integral in the parentheses, and the leftover integral in the single variable $x_n$ is pretty simple.