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Source. Published p. 357, Theorem 20 (published scan).

Statement. The orthogonal product of two finite Ramsey configurations is Ramsey. More precisely, if a finite $T\subseteq\mathbb R^{N_1}$ forces $K_1$ in $r$ colors and $|T|=t$, and $R(K_2,N_2,r^t)$ holds, then $R(K_1\times K_2,N_1+N_2,r)$ holds.

Complete proof. Fix $r$. By compactness, choose a finite witness $T$ for $K_1$ in $r$ colors, with $t=|T|$. Since $K_2$ is Ramsey, choose $N_2$ and a finite witness $S\subseteq\mathbb R^{N_2}$ for $K_2$ in $r^t$ colors.

Given any $r$-coloring $c$ of $T\times S$, color $y\in S$ by its pattern $(c(x,y))_{x\in T}$. There are at most $r^t$ patterns. Thus a copy $K_2'\subseteq S$ has a constant pattern. Choose $y_0\in K_2'$ and color $x\in T$ by $c(x,y_0)$. There is a monochromatic copy $K_1'\subseteq T$. Constancy of the patterns makes $c$ constant on $K_1'\times K_2'$. Its squared pair distances are sums of the corresponding squared factor distances, so it is congruent to $K_1\times K_2$. Restrict arbitrary colorings of $\mathbb R^{N_1+N_2}$ to this finite product to finish. $\square$

Source precision. The last dimension-summary line on p. 357 prints $r^{n_1}$ for the second color count, although the proof uses $r^t$ with $t=|T|$. Ambient dimension does not bound the number of points in this finite witness. The quantified statement above retains the witness size; it does not infer the printed stronger dimension bound. The proof itself already has the correct pattern count.

For the extension to copies using more than one color, see theorem_28. The stronger exponential-density product theorem is compiled separately in theorem_2_2.

Bears on. #174.