TITLE: triangulation n-dimensional cube into exactly n! simplices
QUESTION [10 upvotes]: 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?

REPLY [8 votes]: 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.