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.

Let $X$ be a region in $\mathbb R^n\times\{0\}\subset\mathbb R^{n+1}$, with volume $V$. Now take a point in $\mathbb R^{n+1}$ and form a new solid by connecting this point by lines to all of the points of $X$. This is a "cone on X." Then it is not hard to show, for example by integration, that the $n+1$-dimensional volume of this cone is $\frac{1}{n+1} Vh$, where $h$ is the last coordinate of the cone point. Thus the orthant $x_1+x_2\leq 1,x_1,x_2\geq 0$ has area $1/2$, and in general the orthant $x_1+\cdots+x_n\leq 1, x_i\geq 0$ has volume $1/n$ times the volume of the previous orthant, which establishes the general formula of $1/n!$.
