triangulation n-dimensional cube into exactly n! simplices This question is similar to Find the smallest triangulation of the n-dimensional but easier : 
How to show the n-dimensional cube can be triangulated into
exactly n! simplices?
 A: If the cube has coordinates $x_i$ ($1\leq i\leq n$, $0\leq x_i\leq 1)$ then for every permutation $\sigma\in S_n$ you have the simplex given by $0\leq x_{\sigma(1)}\leq x_{\sigma(2)}\leq\dots\leq  x_{\sigma(n)}\leq 1$.
A: You can triangulate $Δ^n × I$ into $(n + 1)$ $(n + 1)$-simplices ($Δ^n$ is $n$-simplex, $I$ is unit interval $[0, 1]$). See [Hatcher, Proof of 2.10] where it is used to prove homotopy invariance of singular homology. The idea is following: Let $[v_0, …, v_n]$ be the simplex $Δ^n × \{0\}$ and $[w_0, …, w_n] = Δ^n × \{1\}$. Then the triangulation is $\{[v_0, …, v_i, w_i, …, w_n]: i ≤ n\}$.
Now you can triangulate the cube inductively: $I^n = I × \bigcup_{i < (n - 1)!} Δ_i^{n - 1} = \bigcup_{i < (n - 1)!} \bigcup_{j < n} Δ_{ij}^n$.
A: Pick one vertex from your $n$-cube. This vertex has $n$ hyper-faces opposite to it.
Each hyper-face itself is a $(n-1)$-cube and be triangulated into $(n-1)!$ copies of $(n-1)$-simplex. Forming convex hulls with the original vertex gives you $n! = n \times (n-1)!$ copies of $n$-simplex.
