 Solution

True. A useful fact is that squaring is strictly increasing on ordinals. If $A<B$, monotonicity in both arguments gives $A^2\le BA$, and strict increase in the right argument for the positive left factor $B$ gives $BA<B^2$. Hence $A^2<B^2$.

Suppose, for a contradiction, that $\alpha \beta <\beta \alpha$. Again using monotonicity and associativity of ordinal multiplication,This contradicts the hypothesized equality of the outer terms. Reversing the roles excludes $\beta \alpha <\alpha \beta$. If either factor is zero, they already commute. Therefore the commuting squares of ordinals property isThe proof uses monotonicity, which follows directly from the ordered-copy definition; it does not assume a cancellation law for ordinals.