The choice principle $\text{AC}_{\text{WO}}$ proves a large amount of cardinal arithmetic. It's well-known to imply DC, that successor cardinals are regular, and that for all $X$, there is $\lambda$ such that $\aleph(X)=\aleph^*(X)=\lambda^+.$ Furthermore, we have the following:

($\text{ZF + AC}_{\text{WO}}$) If $\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+,$ then $\text{cf}(\lambda)>\text{cf}(\kappa)$ or $\text{cf}(\lambda)=\omega.$

Pf: Fix $\lambda < \aleph(^{\text{cf}(\kappa)} \kappa)$ such that $\omega_1 \le \text{cf}(\lambda) \le \text{cf}(\kappa)$ and a cofinal sequence $\langle \alpha_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Notice that the domination ordering $\le^*$ on $^{\text{cf}(\lambda)} \lambda$ is well-founded, since it has no descending sequences and DC holds. Any well-ordering $\prec$ of $\lambda$ embeds into $(^{\text{cf}(\lambda)} \lambda, \le^*)$ by $\gamma \mapsto (\xi \mapsto \text{ot}({\gamma \downarrow_{\prec}} \cap \alpha_{\xi}, \prec)),$ so $\text{rk}(^{\text{cf}(\lambda)} \lambda, \le^*) \ge \lambda^+.$

We thus have $\lambda^+ < \aleph^*(^{\text{cf}(\lambda)} \lambda) = \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \text{cf}(\kappa)} \kappa) = \aleph(^{\text{cf}(\kappa)} \kappa).$ $\square$

Of course, AC proves this result without the uncountable cofinality hypothesis, which raises the question,

Does $\text{ZF + AC}_{\text{WO}}$ prove that $\aleph(^{\text{cf}(\kappa)} \kappa) = \lambda^+$ for some $\lambda$ with $\text{cf}(\lambda) > \text{cf}(\kappa)?$

Note that a failure of this implication has consistency strength at least a Woodin cardinal, because then $\lambda$ is a singular cardinal with countable cofinal sequence $A,$ and $\text{HOD}(A)$ computes $\lambda^+$ incorrectly.

Of particular interest is the simplest non-trivial case:

Does $\text{ZF + AC}_{\text{WO}}$ prove $\aleph(\mathbb{R})=\Theta \neq \aleph_{\omega+1}?$

A proof from a stronger partition principle like PP or WPP would also be of interest.