Volume of an n-simplex [duplicate]

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 $n$?

What is the probability that a random sequence of $n$ numbers in $[0,1]$ is sorted (lowest to highest)? that is, let $y_1,y_2,\dots,y_n$ be independent uniform random variables in $[0,1]$. Then the probability $P(y_1\leq y_2\leq \dots \leq y_n)=\frac{1}{n!}$.

Let $A=(a_{ij})$ be the matrix such that $a_{ij}=1$ if $i\geq j$ and $a_{ij}=0$ if $i<j$. Then $\det A = 1$ since $A$ is lower-triangular and $a_{ii}=1$ for all $i$. On the other hand, if you take the two sets:

$$S=\{(x_1,\dots,x_n)^T: 0\leq x_i\leq 1: \sum x_i \leq 1\}$$

and $$T=\{(x_1,\dots,x_n)^T: 0\leq x_1\leq x_2\dots \leq x_n\leq 1\}$$

Then $T$ is the image of $S$ under the transformation $A$, that is, $AS=T$. So $$\frac{1}{n!}=\mathrm{vol}(T) = \det A\cdot\mathrm{vol}(S) = \mathrm{vol}(S)$$

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, $$V(T_n(1))=\frac1{n!}.$$

Hint: The general rule is that the $n$-volume of a simplex is $\frac 1n$ times the $ (n-1)$ volume of the base times the height, which you can prove by integration. The height is $1$, so you have a recurrence.

Try using induction. $T_{1}$ is just in the interval $[0,1]$, which has volume (length) $1$. $T_{2}$ is a right triangle with volume (area) $1/2$. Now imagine the case for $n=3$. When we slice the tetrahedron $T_{3}$ at some height $z\in[0,1]$, we get a cross section that looks like $T_{2}$. But as $z$ gets bigger, the cross section gets smaller. Try to convince yourself that in general we have $$ v(T_{n})=\int_{0}^{1}(1-x)^{n-1}v(T_{n-1})\,\mathrm{d}x=\frac{1}{n}v(T_{n-1}) $$ Then, using the fact that $v(T_{1})=1$, we have $v(T_{n})=1/n!$. Note that we scale by $(1-x)^{n-1}$ instead of by $1-x$ because it is the linear dimensions of the $T_{n-1}$ slice that scale by $1-x$, which translates to the $\mathbb{R}^{n-1}$ volume of the slice scaling by $(1-x)^{n-1}$.