This is more subtle than it might first appear. First of all, how do we triangulate $\Delta^n \times \Delta^m$? One answer is to use barycentric subdivision: see Q5 here. However, there is also a triangulation whose vertices are pairs of vertices in the two factors. Either way, once have decided on a coherent procedure for triangulating $\Delta^n \times \Delta^m$, it is a relatively straightforward matter to extend this to a procedure for triangulating products of abstract simplicial complexes in general.
Let me describe the second option in more detail. Let $K$ and $L$ be abstract simplicial complexes and choose a linear ordering of the vertices. We define $K \otimes L$ to be the following abstract simplicial complex:
- Its vertices are pairs $(x, y)$, where $x$ is a vertex of $X$ and $y$ is a vertex of $Y$.
- An $n$-simplex in $K \otimes L$ is a set $\{ (x_0, y_0), \ldots, (x_n, y_n) \}$ such that $x_0 \le \cdots \le x_n$, $y_0 \le \cdots \le y_n$, $\{ x_0, \ldots, x_n \}$ is a simplex in $K$ and $\{ y_0, \ldots, y_n \}$ is a simplex in $L$. (We allow the possibility that the corresponding simplices in $K$ and $L$ are of dimension $< n$, but the sum of their dimensions is always $\ge n$.)
For example, $\Delta^1 \otimes \Delta^1$ corresponds to $⧄$. (If we had neglected the ordering, we would instead get $\Delta^3$!) There are evident simplicial maps $K \otimes L \to K$ and $K \otimes L \to L$, and their geometric realisations induce a homeomorphism $| K \otimes L | \to | K | \times | L |$, as required. (As before, first verify the claim for the standard simplices.)
