Cardinality of the set $\{(\aleph_{\omega_1})^\kappa : 1 \leq \kappa < \aleph_{\omega_1}\}$

Question

I'm trying to figure out a way to answer the following task:

Suppose that for every $ \alpha, \beta < \omega_1 $ there exists a $ \gamma < \omega_1 $ such that $ \aleph_{\alpha}^{\aleph_\beta} = \aleph_\gamma $. Compute the cardinality of the following set $$\{(\aleph_{\omega_1})^\kappa : 1 \leq \kappa < \aleph_{\omega_1}\}$$

Obviously the answer is $ \leq \aleph_{\omega_1} $, but I don't have any great ideas on how to pursue the answer - should I seek the other inequality? How do i use the assumption about exponentiating cardinals?

I would appreciate some hints

Answer 1

Well, it looks as if I'm going to answer my own question.

It's not difficult to prove the following result, knowing Tarski's and Bukovský's identities for exponentiating cardinals:

Suppose $ \lambda < \kappa $. If for all $ \mu < \kappa ~ \mu^\lambda < \kappa$, then $$\kappa^\lambda = \begin{cases}\kappa &\lambda < \operatorname{cf}(\kappa)\\\kappa^{\operatorname{cf}(\kappa)}&\text{otherwise}\end{cases}$$

So what the assumption gives us is that if $ \lambda, \mu < \kappa = \aleph_{\omega_1}$ and by this lemma, this set will contain only two elements: $ \aleph_{\omega_1}, \aleph_{\omega_1}^{\omega_1} $

I would appreciate a verification of this answer