# Does the Cardinal Supremum Commute with the Cardinal Power?

Let $$\kappa,\tau$$ be two cardinals and $$\{\varkappa_\alpha\}_{\alpha<\kappa}$$ an indexed set of cardinals. Is it true that $$\sup_{\alpha<\kappa}(\varkappa_\alpha^\tau)=\left(\sup_{\alpha<\kappa}\varkappa_\alpha\right)^\tau ?$$

I clearly see $$\leq$$. However, I am unsure whether the converse inequality holds generally true. For finite $$\tau$$, the axiom of choice proves this statement directly noting that, for infinite $$\varkappa$$, $$\varkappa^\tau=\varkappa$$. But I'm pretty sure that the statement for finite $$\tau$$ can be proven without the axiom of choice. For infinite $$\tau$$, I also suspect that this holds in $${\sf ZF}$$, but I am not so sure. Thanks in advance!

$$\sup_{n<\omega} (\aleph_n)^\omega \le \sup_{n<\omega} \aleph_{n+1} = \aleph_\omega < \aleph_{\omega+1} = (\aleph_\omega)^\omega = (\sup_{n<\omega} \aleph_n)^\omega$$
As a concrete example, take the sequence $$(\beth_n)_{n<\omega}$$ with $$\tau=\omega$$. It is easy to see that we have $$(\beth_n)^{\omega}=\beth_n$$ for all $$n\geq 1$$. Thus, the left hand side is just $$\beth_\omega$$, which is less than $$(\beth_\omega)^{\omega}$$ by König's Theorem.