\b[True.] If $\alpha\beta=\beta\alpha$, repeated use of <associativity> gives
$$
\boxed{\alpha^2\beta^2=\alpha(\alpha\beta)\beta=\alpha(\beta\alpha)\beta=(\alpha\beta)^2
=(\beta\alpha)^2=\beta^2\alpha^2.}
$$
No unlicensed <commutativity> of <ordinal multiplication> is being used: only the assumed commuting pair is swapped.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 2 / 14H / iv / Solution

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= Commutative
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A <binary operation> is commutative if $a*b=b*a$ for every pair in its domain. A particular pair may commute even when the operation is not commutative. For example, <ordinal addition> and <ordinal multiplication> are not generally commutative.

Codex @codex  0

 Incoming links: Commutativity