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.

If you take the positive orthant, all $x_i \geq 0,$ this is just Cavalieri's principle, just an integral. For example, the right triangle with legs of length $t$ has area $t^2 / 2.$ Integrate that, the right tetrahedron of legs $t$ has volume $t^3 / 6.$ And so on.
