What is the hypervolume of a 4D tetrahedron ($5$-cell)? Here's how this question arose in my mind:


*

*area of a triangle: $\frac{1}{2} \cdot b \cdot h$

*volume of a tetrahedron: $\frac{1}{3} \cdot A \cdot h$

So the 2D object has $\frac{1}{2}$ in the formula, the 3D object has $\frac{1}{3}$ in its formula...does the 4D object have $\frac{1}{4}$ in its formula? And if so, why is there a linear progression instead of something exponential?

The concrete question is, what is the hypervolume of a $4$-dimensional tetrahedron?, aka $5$-cell or pentachoron.
(PS, i saw 2D and 3D as tags, but not 4D or hyper dimensional tag. Don't know what else to tag.)

Indeed, the $n$-simplex = "$n$-dimensional tetrahedron" (and indeed, every $n$-dimensional cone, of which the $n$-simplex is just a special case) has that factor $1/n$. This is easily seen by the fact that the volume of the $n$-dimensional tetrahedron is
\begin{aligned}
V_n &= \int_0^h V_{n-1}(z)\,\mathrm dz\\
&= \int_0^h V_{n-1}(0)\left(\frac{h-z}{h}\right)^{n-1}\,\mathrm dz &&
\text{substitute $w=h-z$}\\
&= \frac{V_{n-1}}{h^{n-1}}\int_0^hw^{n-1}\,\mathrm dw\\
&= \frac{V_{n-1}}{h^{n-1}}\frac{h^n}{n}\\
&= \frac{V_{n-1}h}{n}
\end{aligned}
Here $V_n$ is the volume of the $n$-dimensional cone, $V_{n-1}(z)$ is the $(n-1)$-dimensional volume ("area") of the "horizontal" slice at height $z$ above the base, $V_{n-1}=V_{n-1}(0)$ is the $(n-1)$-dimensional volume of the base, and $h$ is the height of the cone.
In the second line, I've used the fact that all slices of the pace are scaled versions of the base, with a scaling factor $(h-z)/h$, and volumes scale with the power of the dimension.
