Number of faces of Product of Polytopes Is there a simple way to count the number of faces of the product of a polytope with itself n times?
 A: Let us agree to call an $n$-dimensional sub-polytope of some polytope an $n$-face.  Thus we will call edges “1-faces”, and vertices “0-faces”.
Let us write $n_i$ for the number of $i$-faces in a polytope $P$, where $P$ will be clear from context.  For example, for a 3-cube, we have $n_0 = 8, n_1 = 12, n_2 = 6, n_3 = 1,$ and $n_i=0$ for $i$ bigger than 3.
Given integer $k$, and polytope $P$ with its sequence  $n_i$, we want to find $n_2$ for $P^k$.
We can obtain a 2-face of $P^k$ in essentially two ways:


*

*We can select a single 2-face from one of the $k$ copies of $P$, and a single 0-face from each of the $k-1$ other copes, and form the (cartesian) product of these.  This can be done in $$kn_2n_0^{k-1}\tag 1$$ ways.    

*We can select a single 1-face from two of the $k$ copies of $P$, and a single 0-face from each of the $k-2$ other copies, and form the product of these.  This can be done in $$\binom k2 n_1^2n_0^{k-2}\tag 2$$ ways.    
The total number of 2-faces in $P^k$ is the sum of (1) and (2): $$n_0^{k-2}\left(k\cdot n_2n_0+
\binom k2 n_1^2\right)
$$ 
As a check, we now consider the $k$-fold product of the interval $I$.  For $I$ we have $n_0=2, n_1=1, $and $n_i=0$ for $i>1$.  The formula says that $I^k$ has $$2^{k-2}\left(k\cdot0\cdot 2+\binom k2\cdot 1^2\right) = \binom k2 2^{k-2}$$ 2-faces.  This is in fact the number of 2-faces in a $k$-cube for each $k≥0$. (OEIS A001788)

Analogous formulas for the number of $i$-faces can be derived similarly. For example, a 3-face can be obtained either:


*

*As the product of a single 3-face and $k-1$ 0-faces

*As the product of a single 2-face, a single 1-face, and $k-2$ 0-faces

*As the product of three 1-faces and $k-3$ 0-faces.


The formula for the number of 3-faces will be the sum of these:
$$
kn_3n_0^{k-1} + 
k(k-1)n_2n_1n_0^{k-2} +
\binom k3 n_1^3n_0^{k-3}
$$
For $k=3$ and $I$ as above, we get 1, which is correct.

Another way to look at this is to consider the 2-fold product $A\times B$.  Let $a_0, a_1\ldots$ be the number of 0-faces, 1-faces, etc., in $A$, and similarly $b_0, b_1\ldots$ for $B$.  Then the number of $i$-faces in $A\times B$ is $$(A\times B)_i = \sum_{j=0}^i a_ib_{j-i}.$$
For example, this tells us that a triangular prism, which is the product of a triangle ($a_i = \langle 3, 3, 1, 0, \ldots\rangle$) and an interval ($b_i = \langle 2,1,0,\ldots\rangle$) has $a_0b_1 + a_1b_0 = 3 + 6 = 9$ edges and $a_0b_2 + a_1b_1 + a_2b_0 = 0 + 3 + 2 = 5$ faces.
