= 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 $|\kappa\times\kappa|\leq\kappa$, contrary to its being a counterexample; the reverse injection is $\xi\mapsto(\xi,0)$. Hence
$$
\boxed{\aleph_\alpha^2=\aleph_\alpha\quad\text{for every }\alpha.}
$$
A sum of finitely many finite cardinals is finite, so \b[$\aleph_0$ 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 \b[$\aleph_1$ 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 \b[$\aleph_2$ is regular]. In contrast,
$$
\aleph_\omega=\sum_{n<\omega}\aleph_n
$$
is a countable sum of smaller cardinals, with $\aleph_0<\aleph_\omega$. Therefore \b[$\aleph_\omega$ is singular], of <cofinality> $\omega$.
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