= Solution
Put $\mu=\operatorname{cf}(\lambda)$ and choose an increasing cofinal sequence of infinite <cardinals> $\langle\lambda_i:i<\mu\rangle$ below $\lambda$. Since $\lambda$ is a <strong limit cardinal>, $2^{\lambda_i}<\lambda$ for every $i$. Encode a subset $A\subseteq\lambda$ by its restrictions $\langle A\cap\lambda_i:i<\mu\rangle$. Cofinality makes this encoding <injective>, giving
$$
2^\lambda\le\prod_{i<\mu}2^{\lambda_i}\le\lambda^\mu.
$$
The second inequality chooses an injection of each restriction set into $\lambda$. Conversely,
$$
\lambda^\mu\le(2^\lambda)^\mu=2^{\lambda\cdot\mu}=2^\lambda,
$$
since $\mu\le\lambda$ and an infinite cardinal satisfies $\lambda\cdot\mu=\lambda$. The two inequalities prove the <singular strong limit power-set identity>
$$
\boxed{2^\lambda=\lambda^{\operatorname{cf}(\lambda)}}.
$$
This does not assume the <singular cardinals hypothesis> or determine either side as $\lambda^+$. The strong-limit assumption is what bounds each restricted power set.
Back to article page
