I have solved two exercises which asked what kind of ordinals satisfied the following conditions:
$\beta + \alpha =\alpha$ for all $\beta< \alpha$
$\beta\cdot \alpha=\alpha$ for all $\beta < \alpha$
Turns out the first works for $\alpha= \omega^{\xi}$ and the second fo $\alpha=\omega^{\omega^{\xi}}$.
I was wondering: what about exponentiation? I don't think it would be possible, but then how could I prove it?
(1). If $0\ne \alpha=\cup \alpha$ then $\beta^{\alpha}=\cup \{ \beta^{\gamma}: \gamma <\alpha \}.$
(2). Let $a_0=\omega$ and $a_{n+1}=\omega^{(a_n)}$ for $n\in \omega.$ Let $\alpha =\cup \{a_n:n\in \omega\}$.
Confirm that $\beta^{\gamma}<\alpha$ if $\beta, \gamma <\alpha.$ By (1) we obtain $1<\beta <\alpha\implies \beta^{\alpha}=\alpha.$
This $\alpha$ is the least epsilon-number. That is, $\omega^{\alpha}=\alpha.$
BTW in your Q, condition 2, that $\beta<\alpha \implies \beta \cdot \alpha=\alpha,$ cannot hold unless $\alpha =0,$ because of the case $\beta=0.$ What it should say is $0<\beta<\alpha \implies ...$ (etc).