I'm not familiar with the combinatorial proof of Hindman's theorem, but the proof using ultrafilters is quite slick and essentially only relies on the associativity of addition. As such, it also automatically gives the multiplicative version of Hindman's theorem: that for every $F:\mathbb{N}\rightarrow\{1,...,k\}$ there is an infinite $A\subseteq\mathbb{N}$ such that $F$ is homogeneous on the set of products of distinct elements of $A$.
My question is whether the exponential analogue of Hindman's theorem is true:
Is it the case that, for every $F:\mathbb{N}\rightarrow\{1,...,k\}$, there is an infinite $A\subseteq\mathbb{N}$ such that $$\mathsf{FinExp}(A):=\{x_1^{x_2^{.^{.^{.^{x_k}}}}}: x_1,...,x_k\in A\mbox{ distinct}\}$$ is $F$-homogeneous?
Certainly the ultrafilter-based proof breaks down: we can define a version of exponentiation in $\beta\mathbb{N}$, but it isn't a semigroup so we don't get idempotents. (EDIT: And I don't even see how to prove that, for every $F:\mathbb{N}\rightarrow\{0,1\}$, there are $a<b$ such that $\{a,b,a^b,b^a\}$ is $F$-homogeneous.) On the other hand, I don't see how to construct a counterexample (and the numbers get sufficiently big sufficiently quickly that experimentation hasn't been much help either).