 Solution

First derive the standard consequence of König theorem for cardinal numbers that every infinite cardinal $\rho$ satisfiesLet $\theta =\mathrm{cf}(\rho )$ and take cofinal ordinals ${\alpha }_i<\rho$, $i<\theta$. Put ${\rho }_i=\max (1,|{\alpha }_i|)<\rho$. Their sum is $\rho$: it is at least the size of the union of the cofinal ordinals and at most $\rho \cdot \theta =\rho$. Choose the next larger cardinal ${\rho }_i^{+}\le \rho$. König theorem for cardinal numbers givesThis also covers $\rho ={\aleph }_0$, using finite cardinals and their next larger finite cardinals.

Suppose $\lambda ={\kappa }^{\mu }$ with both $\kappa ,\mu$ infinite. Infinite cardinal multiplication givesIf $\mathrm{cf}(\lambda )\le \mu$, monotonicity of exponentiation would contradict the preceding strict inequality for $\rho =\lambda$. Thus $\mathrm{cf}(\lambda )>\mu \ge {\aleph }_0$. This direction does not require GCH.

Conversely assume GCH and $\mathrm{cf}(\lambda )>{\aleph }_0$. Every function $\omega \to \lambda$ has bounded range. For each infinite cardinal $\nu <\lambda$, GCH givesFinite bounded ranges are also covered by the bound ${\aleph }_0^{{\aleph }_0}\le {\aleph }_1\le \lambda$. Counting functions according to an ordinal bound on their range now gives ${\lambda }^{{\aleph }_0}\le \lambda \cdot \lambda =\lambda$. Constant functions give the reverse inequality. Thereforewhich is the required representation with $\kappa =\lambda$, $\mu ={\aleph }_0$. This proves both directions.