The $h$-vector of the (simplicial complex dual to the) permutohedron is given by the sequence of Eulerian numbers.

Example: The (dual of the) 2-dimensional permutohedron is a hexagon, with 6 1-cells, 6 0-cells, and 1 -1-cell. So we compute the $h$-vector by $$\begin{array}{c c} (1,6,6) &= 1 \times (1,2,1)\\ & +4 \times (0,1,1) \\ &+1 \times (0,0,1) \end{array}$$ so that its $h$-vector is $(1,4,1)$.

Meanwhile the $h^*$-vector of the unit hypercube is also given by the Eulerian numbers. This means that the $h$-vector of any unimodular triangulation of the hypercube is the Eulerian numbers, padded with two trailing 0's.

Example: Looking at

we see that a unimodular triangulation of the cube has 6 tetrahedra, 18 triangles, 19 edges, and 8 vertices. So we compute the $h$-vector of this triangulation by $$\begin{array}{c c} (1,8,19,18,6) &= 1 \times (1,4,6,4,1)\\ & +4 \times (0,1,3,3,1) \\ &+1 \times (0,0,1,2,1) \\ &+0 \times (0,0,0,1,1) \\ &+0 \times (0,0,0,0,1) \end{array}$$ so that the $h$-vector is $(1,4,1,0,0)$.

Note: There is a pretty canonical unimodular triangulation of the hypercube whose maximal simplices are $0 \leq x_{w_1} \leq x_{w_2} \leq \cdots \leq x_{w_n} \leq 1$ for all permutations $w = w_1\cdots w_n$.

Question: Is there some deeper geometric connection between a (unimodular) triangulation of the hypercube and the (dual to the) permuotohedron suggested by this numerology?

Bonus question: Suppose $\mathcal{P}$ is a $d$-dimensional polytope with $h$-vector $(c_0,c_1,\ldots,c_d)$. Is there are a simple geometric operation we can perform on $\mathcal{P}$ to produce a $(d+1)$-dimensional polytope $\mathcal{Q}$ with $h$-vector $(c_0,c_1,\ldots,c_d,0)$, i.e., with a 0 padded onto the $h$-vector of $\mathcal{P}$?