= Solution
First derive the standard consequence of <König theorem for cardinal numbers> that every infinite <cardinal> $\rho$ satisfies
$$
\rho^{\operatorname{cf}(\rho)}>\rho.
$$
Let $\theta=\operatorname{cf}(\rho)$ 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> $\rho_i^+\leq\rho$. <König theorem for cardinal numbers> gives
$$
\rho=\sum_{i<\theta}\rho_i<\prod_{i<\theta}\rho_i^+\leq\rho^\theta.
$$
This 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 gives
$$
\lambda^\mu=(\kappa^\mu)^\mu=\kappa^{\mu\cdot\mu}=\lambda.
$$
If $\operatorname{cf}(\lambda)\leq\mu$, monotonicity of exponentiation would contradict the preceding strict inequality for $\rho=\lambda$. Thus $\operatorname{cf}(\lambda)>\mu\geq\aleph_0$. This direction does not require <GCH>.
Conversely assume <GCH> and $\operatorname{cf}(\lambda)>\aleph_0$. Every function $\omega\to\lambda$ has bounded range. For each infinite <cardinal> $\nu<\lambda$, <GCH> gives
$$
\nu^{\aleph_0}\leq(2^\nu)^{\aleph_0}=2^{\nu\cdot\aleph_0}=2^\nu=\nu^+\leq\lambda.
$$
Finite bounded ranges are also covered by the bound $\aleph_0^{\aleph_0}\leq\aleph_1\leq\lambda$. Counting functions according to an <ordinal> bound on their range now gives $\lambda^{\aleph_0}\leq\lambda\cdot\lambda=\lambda$. Constant functions give the reverse inequality. Therefore
$$
\boxed{\lambda=\lambda^{\aleph_0},}
$$
which is the required representation with $\kappa=\lambda$, $\mu=\aleph_0$. This proves both directions.
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