 Solution

True: ordinal multiplication on the left distributes over ordinal addition. If $\alpha =0$ both sides are zero. Otherwise fix $\alpha >0$ and $\beta$, and induct on $\gamma$. At zero the identity follows from $\alpha 0=0$. At a successor, use the recursive ordinal multiplication rule and the associativity just proved:For a nonzero limit ordinal $\lambda$, the ordinals $\beta +\xi$ are cofinal in $\beta +\lambda$. Continuity of ordinal multiplication in its right argument givesFor the last equality, $\alpha \xi$ is increasing and cofinal in $\alpha \lambda$, which is a limit ordinal because $\alpha >0$, and ordinal addition is continuous in its right argument. This establishes the limit step and the identity for all ordinals.