($\text{ZF + AC}_{\text{WO}}$) For any cardinals $\kappa_1, \kappa_2,$ there is $\lambda$ such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+$ and $\text{cf}(\lambda)>\kappa_2.$

Pf: Let $\lambda$ be such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+,$ and fix a cofinal sequence $\langle\gamma_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Choose injections $f_{\alpha}: \alpha \rightarrow \lambda$ for $\alpha<\lambda^+.$ For such $\alpha,$ we recursively define $g_{\alpha}: \text{cf}(\lambda) \rightarrow \lambda$ by setting $g_{\alpha}(\xi) = \min(\lambda \setminus \{g_{\beta}(\xi): \beta \in f^{-1}(\gamma_{\xi})\}).$

Notice that $\alpha \mapsto g_{\alpha}$ injects $\lambda^+$ into $^{\text{cf}(\lambda)} \lambda,$ so $\aleph(^{\kappa_2} \kappa_1)=\lambda^+ < \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \kappa_2} \kappa_1).$ Clearly $\text{cf}(\lambda)> \kappa_2.$ $\square$

Corollary: $\text{AC}_{\text{WO}}$ does prove $\Theta \neq \aleph_{\omega+1}.$

($\text{ZF + AC}_{\text{WO}}$) For any cardinals $\kappa_1, \kappa_2,$ there is $\lambda$ such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+$ and $\text{cf}(\lambda)>\kappa_2.$

Pf: Let $\lambda$ be such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+,$ and fix a cofinal sequence $\langle\gamma_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Choose injections $f_{\alpha}: \alpha \rightarrow \lambda$ for $\alpha<\lambda^+.$ For such $\alpha,$ we recursively define $g_{\alpha}: \text{cf}(\lambda) \rightarrow \lambda$ by setting $g_{\alpha}(\xi) = \min(\lambda \setminus \{g_{\beta}(\xi): \beta \in f_{\alpha}^{-1}(\gamma_{\xi})\}).$

Notice that $\alpha \mapsto g_{\alpha}$ injects $\lambda^+$ into $^{\text{cf}(\lambda)} \lambda,$ so $\aleph(^{\kappa_2} \kappa_1)=\lambda^+ < \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \kappa_2} \kappa_1).$ Clearly $\text{cf}(\lambda)> \kappa_2.$ $\square$

Corollary: $\text{AC}_{\text{WO}}$ does prove $\Theta \neq \aleph_{\omega+1}.$

($\text{ZF + AC}_{\text{WO}}$) For any cardinals $\kappa_1, \kappa_2,$ there is $\lambda$ such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+$ and $\text{cf}(\lambda)>\kappa_2.$

Pf: Let $\lambda$ be such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+,$ and fix a cofinal sequence $\langle\gamma_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Choose injections $f_{\alpha}: \alpha \rightarrow \lambda$ for $\alpha<\lambda^+.$ For such $\alpha,$ we recursively define $g_{\alpha}: \text{cf}(\lambda) \rightarrow \lambda$ by setting $g_{\alpha}(\xi)$ to be least such that for all $\beta \in f^{-1}_{\alpha}(\gamma_{\xi}),$ $g_{\alpha}(\xi) \neq g_{\beta}(\xi).$

Notice that $\alpha \mapsto g_{\alpha}$ injects $\lambda^+$ into $^{\text{cf}(\lambda)} \lambda,$ so $\aleph(^{\kappa_2} \kappa_1)=\lambda^+ < \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \kappa_2} \kappa_1).$ Clearly $\text{cf}(\lambda)> \kappa_2.$ $\square$

Corollary: $\text{AC}_{\text{WO}}$ does prove $\Theta \neq \aleph_{\omega+1}.$

($\text{ZF + AC}_{\text{WO}}$) For any cardinals $\kappa_1, \kappa_2,$ there is $\lambda$ such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+$ and $\text{cf}(\lambda)>\kappa_2.$

Pf: Let $\lambda$ be such that $\aleph(^{\kappa_2}\kappa_1)=\lambda^+,$ and fix a cofinal sequence $\langle\gamma_{\xi}: \xi<\text{cf}(\lambda) \rangle \subset \lambda.$ Choose injections $f_{\alpha}: \alpha \rightarrow \lambda$ for $\alpha<\lambda^+.$ For such $\alpha,$ we recursively define $g_{\alpha}: \text{cf}(\lambda) \rightarrow \lambda$ by setting $g_{\alpha}(\xi) = \min(\lambda \setminus \{g_{\beta}(\xi): \beta \in f^{-1}(\gamma_{\xi})\}).$

Notice that $\alpha \mapsto g_{\alpha}$ injects $\lambda^+$ into $^{\text{cf}(\lambda)} \lambda,$ so $\aleph(^{\kappa_2} \kappa_1)=\lambda^+ < \aleph(^{\text{cf}(\lambda)} \lambda) \le \aleph(^{\text{cf}(\lambda) \cdot \kappa_2} \kappa_1).$ Clearly $\text{cf}(\lambda)> \kappa_2.$ $\square$

Corollary: $\text{AC}_{\text{WO}}$ does prove $\Theta \neq \aleph_{\omega+1}.$