Area of standard simplex The standard $n$-simplex contains all points $\vec{x} \in \mathbb{R}^{n + 1}$ such that $0 \le x_i \le 1$ and $\vec{x} \cdot \vec{1} = 1$
The standard 2-simplex is an equilateral triangle with side length $\sqrt{2}$ and vertices at (1, 0, 0), (0, 1, 0), and (0, 0, 1). The area is $\sqrt{3}/2$.
The standard 1-simplex is a line with vertices at (1, 0) and (0, 1). The length is $\sqrt{2}$
What is the area of the standard $n$-simplex?
Is it $\sqrt{n + 1} / n!$ ?
 A: From this SE-quest you have the height $h_D$ of the unit-sided regular simplex as
$$h_D=\sqrt{\frac{D+1}{2D}}$$
Further you obviously have the dimensional recursion on the volume $V_D$
$$V_D=\frac1D\ V_{D-1}\ h_D$$
With the obvious recursion start of $V_1=1$ you thus get
$$V_D=\prod_{d\leq D}\frac{h_d}d=\frac1{D!}\sqrt{\frac{D+1}{2^D}}$$
(The factor $2^D$ in the denominator surely could be omitted, if you deal with a simplex with sidelength of $\sqrt2$ units instead.)
--- rk
A: I found an answer by integrating over the $n$-simplex moved to $\mathbb{R}^{n}$.
The proof would be much simpler if it is possible to directly integrate over the $n$-simplex, which is in $\mathbb{R}^{n+1}$.
Average of all vertex except first one:
$\vec{a} = \frac{1}{n} \sum_{i = 2}^{n + 1} \vec{e}_i$
Height $\vec{h}$ is between one vertex and the average of the remaining vertex.
$\vec{h} = \vec{a} - \vec{e}_{1}$
Height vector is perpendicular to all vectors in the base:
$\vec{h} \cdot (\vec{a} - \vec{e}_i ) = (\vec{a} - \vec{e}_{1}) \cdot (\vec{a} - \vec{e}_i)$
$ = \vec{a}\cdot\vec{a}
-\vec{a}\cdot\vec{e}_{i}
- \vec{e}_{1}\cdot\vec{a}
+ \vec{e}_{1}\cdot\vec{e}_{i}$
$ = \frac{n + 1}{n} -\frac{1}{n} - 0 + 1 = 0$
Length of height vector is:
$l_{n} = \sqrt{\vec{h}\cdot\vec{h}} = \sqrt{\vec{a}\cdot \vec{a} + \vec{e}_{1}\cdot \vec{e}_{1}} = \sqrt{\left(\frac{1}{n^{2}}\sum_{i=2}^{n+1} \vec{e}_{i}\cdot\vec{e}_{i} \right ) + 1}=\sqrt{\frac{1}{n} + 1}=\sqrt{\frac{n + 1}{n}}$
Let $A_{n}(s)$ be area of the standard $n$-simplex scaled by $s$ along each of all axes.
$A_{1} = \sqrt{2}s$
Integrate along the height vector for the scaled standard $(n + 1)$-simplex.
$A_{n+1}(s) = \int_{0}^{sl_{n+1}} A_{n}\left (\frac{sx}{sl_{n+1}} \right)dx = sl_{n+1}\int_{0}^{1} A_{n}(su)du$
$A_{2}(s) = sl_{2}\int_{0}^{1} \sqrt{2}sudu = l_{2} \sqrt{2} \frac{1}{2}s^{2}= \frac{\sqrt{3}}{2}s^{2}$
$A_{3}(s) =  sl_{3}\int_{0}^{1} sl_{2}\int_{0}^{1} \sqrt{2}sudu$
$$A_{n}(s) =  s^{n} \left( \prod_{1}^{n}l_{n} \right) \int_{0}^{1} \dots \int_{0}^{1} udu$$
$$A_{n}(s) =  \frac{ s^{n} \sqrt{n+1}}{n!}$$
