$\kappa <\operatorname{cf}(2^\kappa)$ without König's inequality

Question

How can I prove $\kappa<\operatorname{cf}(2^\kappa)$ inequality without using König's inequality?
We got this as a practice exercise, but I don't know how to approach this without König.
Any hint would be awesome!

I tried it with different other Corollaries which I found in Jech, but nothing worth to post it here.

Answer 1

This follows from $\kappa \lt \kappa^{cf(\kappa)}$ that can be proved without Konig's theorem (see e.g. Jech/Set Theory Theorem 3.11).

Now to get the requested equality substitute $2^\kappa$ instead of $\kappa$ and use the fact that $\kappa \le cf(2^\kappa)$ (that can be seen by assuming the opposite, and getting a contradiction from $2^{k * cf(2^\kappa)} \lt 2^{\kappa * \kappa}$).