Show that $\aleph_\alpha^{\aleph_\beta} = \aleph_\alpha$ when $\aleph_\beta<\operatorname{cf}(\aleph_\alpha)$ and some additional conditions are true.
I have been reading the book "Introduction to Set Theory" by Jech and Hrbacek. I have come to an exercise that I have a question about in the chapter on the arithmeric of cardinal numbers in the section on cardinal exponentiation. The exercise statement goes : $\newcommand{\cf}{\operatorname{cf}}$
Let $\alpha$ be a limit ordinal and let $\aleph_{\beta} < \cf(\aleph_{\alpha})$. Show that if $\aleph_{\xi}^{\aleph_{\beta}} \leq \aleph_{\alpha}$ for all $\xi < \alpha$, then $\aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\alpha}$. [Hint : If $X \subseteq \omega_{\alpha}$ is such that $|X| = \aleph_{\beta}$, then $X \subseteq \omega_{\xi}$ for some $\xi < \alpha$.]
My current solution uses the folowing results from the book :
Theorem 9.3.10 :
Let us assume the Generalized Continuum Hypothesis. If $\aleph_{\alpha}$ is a singular cardinal, then :
\begin{equation}
\aleph_{\alpha}^{\aleph_{\beta}} =
\begin{cases}
\aleph_{\alpha} & \text{ if } \aleph_{\beta} < \cf(\aleph_{\alpha})\\
\aleph_{\alpha + 1} & \text{ if } \cf(\aleph_{\alpha}) \leq \aleph_{\beta} \leq \aleph_{\alpha}\\
\aleph_{\beta + 1} & \text{ if } \aleph_{\beta} \geq \aleph_{\alpha}
\end{cases}
\end{equation}
Theorem 9.3.8 :
Let us assume the Generalized Continuum Hypothesis. If $\aleph_{\alpha}$ is a regular cardinal, then :
\begin{equation}
\aleph_{\alpha}^{\aleph_{\beta}} =
\begin{cases}
\aleph_{\alpha} & \text{ if } \beta < \alpha \\
\aleph_{\beta + 1} & \text{ if } \beta \geq \alpha
\end{cases}
\end{equation}
Lemma 9.2.8:
We have for all limit ordinals $\alpha$ :
\begin{equation}
\cf(\alpha) = \alpha \iff \alpha \ \text{is a regular cardinal }
\end{equation}
My issue:
My current proof shows that, whenever $\alpha$ is a limit
cardinal and $\aleph_{\beta} < \cf(\aleph_{\alpha})$, then $\aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\alpha}$. It does not depend on the entirety of the stated premise being true. This suggests that there is a problem with the solution that I currently have. My solution so far is given below :
If $\aleph_{\alpha}$ is a singular cardinal then by theorem 9.3.10 : \begin{equation} \aleph_{\beta} < \cf(\aleph_{\alpha}) \Rightarrow \aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\alpha} \; \checkmark \end{equation} What if $\aleph_{\alpha}$ is regular ? We have by theorem 9.3.8 : \begin{equation} \beta < \alpha \Rightarrow \aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\alpha} \end{equation} So need to prove : \begin{equation} \beta < \alpha \end{equation} We know by lemma 9.2.8 : \begin{equation} \aleph_{\alpha} \text{ is regular } \Rightarrow \cf(\aleph_{\alpha}) = \aleph_{\alpha} \end{equation} So : \begin{equation} \aleph_{\beta} < \cf(\aleph_{\alpha}) \Rightarrow \aleph_{\beta} < \aleph_{\alpha} \end{equation} Where $\aleph_{\alpha}$ is regular. Why can't I just say : \begin{equation} \aleph_{\beta} < \aleph_{\alpha} \Rightarrow \beta < \alpha \; \checkmark \end{equation} and therefore : \begin{equation} \aleph_{\alpha}^{\aleph_{\beta}} = \aleph_{\alpha} \; \checkmark \end{equation}
Can someone help with this ?