Volume of $n$-simplex in $\Bbb R^m$ without Cayley-Menger determinant The $n$-volume of an irregular $n$-simplex in $\mathbb{R}^m$ with $m\gt n$ can be calculated by the Cayley-Menger determinant using squared edge lengths. Are there alternative approaches to calculate the $n$-volume if $m \gt n$ without using a Cayley-Menger determinant? Reshaped matrices (like here) that are in principal equal variants of the Cayley-Menger determinant  do not count.
 A: Here's one alternative method.
If $\ V_n\big(v_0,v_1,\dots,v_n\big)\ $ is the $n$-volume of the $n$-simplex $\ \mathcal{S}\ $with vertices $\ v_0,v_1,\dots,v_n\ $, then
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)\big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|}{n}\ ,
$$
where $\ P_n\ $ is the perpendicular projection onto the $\ n-1$-dimensional space spanned by $\ v_{n-1}-v_0,$$v_{n-2}-v_0,$$\dots, v_1-v_0\ $.  The factor $\ \big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|\ $ in this product is just the height of the vertex $\ v_n\ $ above the $n-1$-simplex with vertices $\ v_0,v_1,\dots,v_{n-1}\ $, taken as a base of $\ \mathcal{S}\ $. By induction, therefore,
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{1}{n!}\|v_1-v_0\|\prod_{j=1}^{n-1}\big\|\big(I-P_j\big)\big(v_{j+1}-v_0\big)\big\|\ .
$$
Derivation of the formula
If $\ h_n=\big\|\big(I-P_{n-1}\big)\big(v_n-v_0\big)\big\|\ $ is the height of the vertex $\ v_n\ $ of the simplex $\ \mathcal{S}\ $ above its base, $\ \text{Conv}\big(v_0,v_1,\dots,v_{n-1}\big)\ $, and $\ A_{n-1}(x)\ $ is the $\ n-1$-volume of the cross-section of  $\ \mathcal{S}\ $ at height $\ x\ $ above that base, then the volume of $\ \mathcal{S}\ $ is
$$
\int_0^{h_n}A_{n-1}(x)dx\ .
$$
But the cross-section of $\ S\ $ at height $\ x\ $ above the base is the $\ n-1$-simplex with vertices $\ v_0(x),v_1(x), \dots, v_{n-1}(x)\ $, where
$
v_i(x)=\left(1-\frac{x}{h_n}\right)v_i+\left(\frac{x}{h_n}\right)v_n
$. This simplex is similar to the base $\ \text{Conv}\big(v_0,v_1,\dots,v_{n-1}\big)\ $, scaled down by a linear factor $\ \frac{x}{h_n}\ $ $\big($since $\ v_i(x)-v_j(x)=\left(\frac{x}{h_n}\right)\big(v_i-v_j\big)\ $$\big)$. Its $n-1$-volume $\ A_{n-1}(x)\ $is therefore $\ \left(\frac{x}{h_n}\right)^{n-1}V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)\ $. The volume of $\ \mathcal{S}\ $ is therefore
\begin{align}
\int_0^{h_n}A_{n-1}(x)dx&=\int_0^{h_n}\left(\frac{x}{h_n}\right)^{n-1}V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)dx\\
&=\frac{h_n\,V_{n-1}\big(v_0,v_1,\dots,v_{n-1}\big)}{n}\ ,
\end{align}
as stated above.  The formula is a generalisation of the instances that the area of a triangle is half the length of its base multiplied by its height, and the volume of a tetrahedron is one third of the area of its base multiplied by its height.
Simplification
The above formula can be written more simply if the Gram-Schmidt procedure is used to orthonormalise the vectors $\ v_1-v_0,v_2-v_0,\dots, v_n-v_0\ $:
\begin{align}
u_1&=\frac{v_1-v_0}{\big\|v_1-v_0\big\|}\ ,\\
u_j&=\frac{v_j-v_0 -\sum_\limits{i=1}^{j-1}\left\langle v_j-v_0,u_i\right\rangle}{\big\|v_j-v_0 -\sum_\limits{i=1}^{j-1}\left\langle v_j-v_0,u_i\right\rangle\big\|}
\end{align}
for $\ j=2,3,\dots,n\ $. Then $\ \big(I-P_j\big)\big(v_{j+1}-v_0\big)=$$\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle u_{j+1}\ $, and $\ \big\|\big(I-P_j\big)\big(v_{j+1}-v_0\big)\big\|=$$\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle\ $. Therefore
$$
V_n\big(v_0,v_1,\dots,v_n\big)=\frac{\prod_\limits{j=0}^{n-1}\left\langle\big(v_{j+1}-v_0\big),u_{j+1}\right\rangle}{n!}\ .
$$
Postscript
It's occurred to me that the orthonormalisation, $\ u_1,u_2,\dots,u_n\ $, of $\ v_1-v_0,v_2-v_0,\dots, v_n-v_0\ $ allows one to define an isometry $\ v_0+\sum_\limits{i=1}^nx_iu_i\mapsto\big(x_1,x_2,\dots,x_n\big)\ $ from the affine hull of $\ \mathcal{S}\ $ onto $\ \mathbb{R}^n\ $.  The image of $\ \mathcal{S}\ $ under this isometry will be an $n$-simplex in $\ \mathbb{R}^n\ $  congruent to $\ \mathcal{S}\ $, and therefore having the same volume.  Therefore any of the well-known formulas for the volume of an $n$-simplex in $\ \mathbb{R}^n\ $ will give another method for calculating the volume of $\ \mathcal{S}\ $.
