# If $\lambda$ is a limit cardinal then $\kappa^\lambda = (\kappa^{<\lambda})^{cf \lambda}$ for each cardinal $\kappa$.

I'm reading Thomas Jech's "Set Theory" and there is a theorem about properties of continuum function and cardinals:

Theorem 5.16:

1. If $$\kappa < \lambda$$ then $$2^\kappa \leq 2^\lambda$$.

2. $$\operatorname{cf} 2^\kappa > \kappa$$.

3. If $$\kappa$$ is a limit cardinal then $$2^\kappa = (2^{< \kappa})^{\operatorname{cf} \kappa}$$.

The proof of (3) is following: Let $$\kappa = \sum_{i < \operatorname{cf} \kappa} \kappa_i$$ where $$\kappa_i < \kappa$$. Then $$2^\kappa = 2^{\sum_{i < \operatorname{cf} \kappa} \kappa_i} = \prod_{i < \operatorname{cf} \kappa} 2^{\kappa_i} \leq \prod_{i < \operatorname{cf} \kappa} 2^{< \kappa} = (2^{< \kappa})^{\operatorname{cf} \kappa} \leq (2^\kappa)^{\operatorname{cf} \kappa} = 2^\kappa$$

The proof relies only on properties $$\kappa < \lambda$$ implies $$2^\kappa \leq 2^\lambda$$, $$\kappa < \lambda$$ implies $$\kappa^\mu \leq \lambda^\mu$$, and connections between infinite sums and products of cardinals. But all such properties are still satisfied if we replace $$2$$ by some arbitrary cardinal $$\kappa$$. So, following same proof, we can deduce a stronger proposition: If $$\lambda$$ is a limit cardinal then $$\kappa^\lambda = (\kappa^{<\lambda})^{cf \lambda}$$ for each cardinal $$\kappa$$.

Am I missing something? I was given such problem as an exercise, and it seems so simple that I started to doubt if I'm getting everything right.

Thanks!

No, you aren't missing anything. (And note also that $$\lambda$$ doesn't even need to be a limit cardinal for the formula to hold, just an infinite cardinal, although the case where it is a successor cardinal is less interesting.)