Operations on ordinal numbers

The result is false. A counterexample is obtained by taking $w=\omega^2+\omega+1$.

Note $w+w=\omega^2\cdot2+\omega+1$. Now, adding $z=\omega^2\cdot2+\omega+1$ to itself $\omega$ times is $\omega^3$, and adding $z$ to itself $\omega^2$ times is therefore $\omega^4$.

Then $(w+w)w=z(\omega^2+\omega+1)=\omega^4+\omega^3+\omega^2\cdot2+\omega+1$.

On the other hand, $ww=\omega^4+\omega^3+\omega^2+\omega+1<(w+w)w$.

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In general, one can obtain many counterexamples by considering $w$ whose Cantor normal form includes indecomposables of several kinds, not all of them limit ordinals.