Triangulation of hypercubes into simplices A square can be divided into two triangles. A 3-dimensional cube can be divided into 6 tetrahedrons. Into what number of simplices an n-dimensional hypercube can be divided? (For example, a 4-hypercube, or a 5-hypercube.)

Noting that the volume of a cone-like object
$$O = \bigcup_{t\in[0,h]} \{t\}\times \frac thA$$
Is given by $\mathrm{Vol}^n(O) = \frac hn \mathrm{Vol}^{n-1}(A)$ suggests that we need $n$ $(n-1)$-simplices to get from a $n-1$-hypercube (hyper-parallelepiped) to a $n$-hypercube, i.e.
$$N(n) = n \cdot N(n-1); \quad N(1) = 1 \Rightarrow N(n) = n!$$
Note that this only constitutes an upper bound. For 3D it gives $6$ when there is in fact a triangulation with only $5$ simplices.
