 Solution

Work with well-ordered cardinals, as usual with the axiom of choice. We first prove the infinite cardinal arithmetic identity ${\kappa }^2=\kappa$. If a least infinite counterexample $\kappa$ existed, order pairs of ordinals below $\kappa$ first by their maximum coordinate, then lexicographically within each level. Every initial segment ending at $(\xi ,\eta )$ lies inside $(\max (\xi ,\eta )+1{)}^2$. Its cardinal is less than $\kappa$, by the smaller-cardinal identity or finite counting. This well-order cannot have an element with $\kappa$ predecessors, so its order type is at most the initial ordinal $\kappa$. Thus $|\kappa \times \kappa |\le \kappa$, contrary to its being a counterexample; the reverse injection is $\xi \mapsto (\xi ,0)$. HenceA sum of finitely many finite cardinals is finite, so ${\aleph }_0$ is regular. A sum of fewer than ${\aleph }_1$ smaller cardinals is a countable union of countable sets and has size at most ${\aleph }_0^2={\aleph }_0$, so ${\aleph }_1$ is regular.

Likewise fewer than ${\aleph }_2$ summands means at most ${\aleph }_1$ summands, each of size at most ${\aleph }_1$. Their sum is at most ${\aleph }_1^2={\aleph }_1<{\aleph }_2$. Thus ${\aleph }_2$ is regular. In contrast,is a countable sum of smaller cardinals, with ${\aleph }_0<{\aleph }_{\omega }$. Therefore ${\aleph }_{\omega }$ is singular, of cofinality $\omega$.