Assuming the Axiom of Choice: $\kappa <{\kappa }^{\mathrm{cf}(\kappa )}$ for every infinite cardinal $\kappa$, and $\mathrm{cf}(2^{\kappa })>\kappa$; in particular $\mathrm{cf}(2^{{\mathrm{\aleph }}_0})>{\mathrm{\aleph }}_0$

Assume the Axiom of Choice (The Axiom of Choice). Let $\kappa$ be an infinite cardinal (Cardinal (initial ordinal) and cardinality). Then:

(a) $\kappa <{\kappa }^{\mathrm{cf}(\kappa )}$ (Cardinal sum $\kappa \oplus \lambda$, product $\kappa \otimes \lambda$ and exponentiation ${\kappa }^{\lambda }$, and why they are written apart from the ordinal operations, Cofinality $\mathrm{cf}(\alpha )$, and regular and singular cardinals);

(b) $\mathrm{cf}(2^{\kappa })>\kappa$;

(c) in particular $\mathrm{cf}(2^{{\mathrm{\aleph }}_0})>{\mathrm{\aleph }}_0$ (The successor cardinal ${\kappa }^{+}$, the alephs ${\mathrm{\aleph }}_{\alpha }$, the beths ${\mathrm{\beth }}_{\alpha }$, successor and limit cardinals, and the identifications ${\mathrm{\aleph }}_0=\omega$ and ${\mathrm{\aleph }}_1={\omega }_1$).

Clause (c) is a genuine restriction on the continuum, and it is proved in ZFC rather than quoted. It rules out every value of $2^{{\mathrm{\aleph }}_0}$ whose cofinality is ${\mathrm{\aleph }}_0$, and it selects none.

What clause (b) rules out, concretely. If $2^{{\mathrm{\aleph }}_0}$ were ${\mathrm{\aleph }}_{\omega }$ then its cofinality would be ${\mathrm{\aleph }}_0$ by ${\mathrm{\aleph }}_0$ is regular in ZF; assuming the Axiom of Choice every successor aleph ${\mathrm{\aleph }}_{\alpha +1}$ is regular; $\mathrm{cf}({\mathrm{\aleph }}_{\omega })={\mathrm{\aleph }}_0$, so ${\mathrm{\aleph }}_{\omega }$ is singular, and under choice it is the least singular infinite cardinal, contradicting clause (c); that refutation is carried out in FALSE: $2^{{\mathrm{\aleph }}_0}={\mathrm{\aleph }}_{\omega }$. The same test applies to any proposed value whose cofinality can be computed. Clause (c) is a restriction and not a determination: it excludes values and selects none.

Why the cofinality, and not the size, is the obstruction. Clause (a) says a cardinal is strictly smaller than itself raised to its own cofinality. Read contrapositively, a cardinal $\mu$ that is a power $2^{\kappa }$ cannot have its cofinality drop to or below $\kappa$, because raising $\mu$ to that exponent would not increase it. The whole argument is the interaction of two facts, König's inequality and the second exponent law, and the second is where Hessenberg: $\kappa \otimes \kappa =\kappa$ for every infinite cardinal $\kappa$, proved in ZF from the canonical well-order of $\kappa \times \kappa$ enters.

Where the Axiom of Choice is spent here. Three times: in the definitions of $\sum$, $\prod$ and $2^{\kappa }$; in step 2.1, to select a bijection $g(\xi )\to \mid g(\xi )\mid$ for every $\xi$ at once; and inside König's theorem: assuming the Axiom of Choice, if ${\kappa }_i<{\lambda }_i$ for every $i\in I$ then ${\sum }_{i\in I}{\kappa }_i<{\prod }_{i\in I}{\lambda }_i$ itself. None of the three is removable by a canonical construction, which is why the whole corollary carries the hypothesis in its statement.