Solution
= Solution
Always true. Under the synthetic definition of <ordinal multiplication>, $\alpha\beta$ is the order type of $\beta$ consecutive copies of $\alpha$. Grouping the copies first into $\gamma$ blocks of $\beta$ copies, or first into single blocks of type $\alpha\beta$, gives the same lexicographic well-order. Hence
$$
\boxed{\alpha(\beta\gamma)=(\alpha\beta)\gamma.}
$$