Why is the permutohedron simple? I am working with the permutohedron in $\mathbb{R}^n$ which is defined as the convex hull of $n!$ vectors as follows: 
$$\Pi_n := conv\{(\sigma(1), \ldots, \sigma(n))\ |\ \sigma \text{ permutation of } [n]\}$$
I want to show that this polytope is simple and $n-1$ dimensional. To prove that $dim(\Pi_n)= n-1$ was easy. I further proved that $\Pi_n$ has exactly $n!$ vertices as expected. 
But why is the permutohedron simple, i.e. why has every vertex exactly $n-1$ neighbors?
I've read that the vertices form an edge if they differ in a transposition between two elements that differ by 1. Unfortunately, I was not able to prove that (this would yield the simplicity). How does one prove that?
Thanks a lot
P.S.: I already checked this thread, however it hasn't received any answers yet.

By @anomaly's comment, you only need to count the edges at any one vertex. Notice further that the projection which simply drops the $n$th coordinate is injective and dimension-preserving on the vertices of the permutahedron, so the convex hull in the first $n-1$ dimensions has the same combinatorial structure as the original permutahedron. In this structure, the set of points with greatest ($n-1$)st coordinate is clearly both (a) on the convex hull and (b) a copy of the order $n-1$ permutahedron. So by induction each vertex of it has $n-2$ edges in that same layer; we just have to find at least one vertex of it that has only a single edge in which the ($n-1$)st coordinate changes. But in particular, looking along the $n-1$st coordinate, the vertices which consist of all permutations of (1,2,...,$n-2$) followed by $n$ lie directly "above" the same permutation of (1,2,...,$n-2$) followed by $n-1$, so they must each have a single additional edge in the overall convex hull, parallel to the $n-1$st axis. This completes the proof.
