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


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*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.)

The $\frac{1}{n}$ term was understood well before calculus.
Archimedes found the volume of a pyramid by slicing it into thin triangles and summing their areas and throwing away the small terms, prefiguring calculus by 2000 years.
In general, it was well known that:
$$\sum^N_{i=0} i^n \sim \frac{N^{n+1}}{n+1}$$
The area under the curve $x^n$ was shown geometrically to be equal to $\frac{x^{n+1}}{n+1}$ for positive integer values of $n$ a few decades before calculus.
However, the concept of higher dimensional geometry only came some 100 years after calculus.
