volume of "$n$-hedron" In $\mathbb{R}^n$, why does the "$n$-hedron" $|x_1|+|x_2|+\dots+|x_n| \le 1$ have volume $\cfrac{2^n}{n!}$? I came across this fact in some of Minkowski's proofs in the field of geometry of numbers.
Thank you.

The $2^n$ comes because you have that many copies of the simplex $0 \le x_i \le 1$
The $n!$ comes from integrating up the volume.  The area of the right triangle is $\frac 12$, the volume of the tetrahedron is $\frac 12 \cdot \frac 13$ and so on.
