Can a set contain the cartesian product with itself? Say $A$ is a a non-empty set, is it possible to have $A^n \times A^m \in A$? For any integers $n,m$,
I'm not asking if this is true in general, I'm only asking if it can be the case for some particular set. If this is true, could you provide an example? If this is always false could you provide a proof? Thank you.

The answer depends on the underlying set theory and the actual symbol under consideration, whether $\in$ or $\subseteq$.
In standard (ZF) set theory, the axiom of foundation prevents the existence of any set as specified. The reason is that sets have a rank, and the rank of any member of a set $A$ is strictly smaller than that of $A$. However, the rank of powers of $A$ and of Cartesian products is at least as large as that of $A$.
There are, however, set theories (such as ZFA, where the A stands for Aczel's anti-foundation axiom) where there are sets $\Omega$ such that $\Omega=\{\Omega\}$. For any such $\Omega$ and the standard definition of ordered pair, $$(\Omega,\Omega)=\{\{\Omega\},\{\Omega,\Omega\}\}=\{\Omega,\Omega\}=\Omega$$ and so all finite powers and all Cartesian products of finite powers of $\Omega$ just coincide with $\Omega$ itself, and for any $n,m$ positive integers, $\Omega^n\times\Omega^m=\Omega\in\Omega$. The anti-foundation axiom also implies the existence of sets $A$ with $A^n\times A^m\in A$ even if $n$ or $m$ is 0. It also implies the existence of such sets if powers are taken in the sense of functions rather than Cartesian products (i.e., for instance., if $A^2$ is understood as the set of functions $f:\{0,1\}\to A$ rather than as the product $A\times A$).
The existence of sets $A$ such that $A^n\times A^m\subset A$ is less problematic (the axiom of foundation plays no role here). It is easy to give examples: the empty set, for instance. Or $V_\omega=\bigcup_k\mathcal P^k(\emptyset)$. In general, given any set $A$, you can easily find a larger set $A'$ with $(A')^n\times(A')^m\subseteq A'$ for any $n,m$: just build $A'$ by a recursive process: $A'=\bigcup_k A_k$, where $A_0=A$, and $$A_{s+1}=A_s\cup\bigcup_{n,m}A_s^n\times A_s^m.$$ (The set $A'$ so constructed in the smallest set containing $A$ and satisfying the desired property.)
