For every $n < \omega$, $\aleph_n^{\aleph_0} = \max(\aleph_n,\aleph_0^{\aleph_0})$ I have proved that if $\aleph_n \leq \aleph_0^{\aleph_0}$, then $\aleph_n^{\aleph_0} \leq \max(\aleph_n,\aleph_0^{\aleph_0})$. Clearly $\aleph_n \leq \aleph_n^{\aleph_0}$ and $\aleph_0^{\aleph_0} \leq \aleph_n^{\aleph_0}$. Therefore, in order to establish the desired equality, I only need to show that if $\aleph_0^{\aleph_0} \leq \aleph_n$, then $\aleph_n^{\aleph_0} \leq \max(\aleph_n,\aleph_0^{\aleph_0})$.
I guess that might be easy, but I just don't see it. Any comments are welcome.
 A: Suppose that there is an $n$ such that $\aleph_0^{\aleph_0}\le\aleph_n$ and $\aleph_n^{\aleph_0}>\aleph_n$, and let $m$ be the least such $n$; clearly $m>0$. Now consider a function $\varphi:\aleph_0\to\aleph_m$; $\aleph_m$ is an uncountable regular cardinal, so $\sup\{\varphi(k):k\in\aleph_0\}<\aleph_m$, and $\varphi$ actually maps $\aleph_0$ into $\eta$ for some ordinal $\eta<\aleph_m$. Writing $^AB$ for the set of functions from $A$ into $B$, we have $$^{\aleph_0}\aleph_m=\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}$$ and hence $$\left|^{\aleph_0}\aleph_m\right|=\left|\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}\right|\;.$$
For each $\eta<\aleph_m$, $|\eta|\le\aleph_{m-1}$, so $$\aleph_m^{\aleph_0}=\left|^{\aleph_0}\aleph_m\right|=\left|\bigcup_{\eta<\aleph_m}{^{\aleph_0}\eta}\right|\le\aleph_m\cdot\aleph_{m-1}^{\aleph_0},$$ and $\aleph_{m-1}^{\aleph_0}\le\aleph_{m-1}$ by the minimality of $m$, so $$\aleph_m^{\aleph_0}\le\aleph_m\cdot\aleph_{m-1}^{\aleph_0}\le\aleph_m\cdot\aleph_{m-1}=\aleph_m\;,$$
contradicting the choice of $m$. Thus, $\aleph_m^{\aleph_0}\le\aleph_m=\max\{\aleph_m,\aleph_0^{\aleph_0}\}$ whenever $\aleph_0^{\aleph_0}\le\aleph_m$.
