A question on cardinal arithmetic

Question

The following problem is exercise 6.12 of Thomas Jech's "Set Theory" and i'd be thankful for some hints:

First we define $ \kappa^{\lt \kappa} $ and $ 2^{\lt \kappa} $:

$ \kappa^{\lt \kappa} = \lim_{\alpha \rightarrow \kappa} \kappa^\alpha $
$ 2^{\lt \kappa} = \lim_{\alpha \rightarrow \kappa} 2^\alpha $

The statement: If $\kappa$ is a regular and limit cardinal, then $\kappa^{\lt \kappa} = 2^{\lt \kappa}$.

Edit I:
The $\kappa^{\lt \kappa} \ge 2^{\lt \kappa}$ part is quite easy since every $\kappa^\alpha$ is greater than or equal to $2^\alpha$ then the suprema must be greater than or equal to, aswell. The tricky part is proving $\kappa^{\lt \kappa} \le 2^{\lt \kappa}$.

Answer 1

Notation: $^d e$ is the set of functions from $d$ to $e$.

Depending on your choice of definitions, $\omega$ could be a regular limit cardinal. In this Q, $\kappa=\omega$ is a trivial case, so assume $\kappa$ is uncountable.

If $a< k$ and $f\in ^a \kappa$ then $f\in ^a b$ for some $b<\kappa.$ For example if $b=\sup \{f(x):x\in a\}+1$ then $b<\kappa$ because (i) $\kappa$ is a limit ordinal, and (ii) $cf(\kappa)=\kappa>|a|\geq|\{f(x):x\in a\}|.$

So $\kappa^{<\kappa}=|\cup \{^a \kappa:a\in \kappa\}|=|\cup \{^a b:a,b\in \kappa\}|.$

If $c$ is an infinite cardinal ordinal then $c^c=2^c.$

For $a,b \in \kappa$ let $c=\max (|a|,|b|,\omega).$ Then $c< k$ and $|^a b|=|b|^{|a|}\leq c^c=2^c.$

The rest should be easy.