TITLE: Volume of $T_n=\{x_i\ge0:x_1+\cdots+x_n\le1\}$
QUESTION [14 upvotes]: Let $T_n=\{x_i\ge0:x_1+\cdots+x_n\le1\}$. I know $T_n$ is tetrahedron.
My question: How can I compute the volume of $T_n$ for every $n$?

REPLY [13 votes]: First suppose that we want to find the volume of $T_n(a)$, then by change of variables $X=aU$, you can see that we have $V(T_n(a))=\color{red}{a^n}V(T_n(1))$.
Since $x_1+\cdots+x_n\le1$ if and only if $x_n\le1$ and $x_1+\cdots+x_{n-1}\le 1-x_n$, we have
$$\begin{align}V(T_n(1))&=\int_{x_n\le 1}\left(\int_{x_1+\cdots+x_{n-1}\le1-x_n}dx_1\cdots dx_{n-1}\right)dx_n\\
&=V(T_{n-1}(1))\int_{x_n\le1}\color{red}{(1-x_n)^{n-1}}dx_n=\frac1 nV(T_{n-1}(1))\end{align}$$
The numbers $V(T_n(1))$ satisfy in the above recursion formula, so
$$V(T_n(1))=\frac1{n!}.$$
One another way is to consider the following integral
$$I=\int_{T_n(a)}e^{-(x_1+\cdots+x_n)}dx_1\cdots dx_n$$
Since $V(T_n(a))=a^nV(T_n(1))$,
$$I=\int_0^\infty e^{-a}dV(T_n(a))=V(T_n(1))\int_0^\infty n a^{n-1}e^{-a}da=n!V(T_n(1))$$
But also we have
$$I=\left(\int_0^\infty e^{-x}dx\right)^n=1$$
Hence, 
$$V(T_n(1))=\frac1{n!}.$$