$\alpha > \omega$ is an $\epsilon$-number iff $\beta^\gamma < \alpha, \forall \beta, \gamma < \alpha$ Prove: $\alpha > \omega$ is an $\epsilon$-number iff $\beta^\gamma < \alpha,   \forall \beta, \gamma < \alpha$.
From left to right:
if $\omega >\beta:  \alpha=\omega^\alpha> \beta^\gamma$
if $\omega =\beta:  \alpha=\omega^\alpha> \omega^\gamma=\beta^\gamma$
if $\omega <\beta:  \alpha=\omega^\alpha> \omega^\gamma$ and then?
Can anyone full prove?
 A: $(\Rightarrow)$ Let $\alpha>\omega$ be an $\epsilon$-number. For any ordinal $\beta$ let us define the degree of $\beta$, $\deg(\beta)$ the highest $\gamma$ that is an exponent of the Cantor normal form of $\beta$; so for instance an ordinal $\beta$ is an $\epsilon$-number if and only of $\deg(\beta)=\beta$.
Let us prove first $\forall \beta,\gamma<\alpha[\beta+\gamma<\alpha]$. As $\alpha=\omega^\alpha$, it follows that $\deg(\beta),\deg(\gamma)<\alpha,$ however, beacuse of this, $\deg(\beta+\gamma)=\max(\deg(\beta),\deg(\gamma))<\alpha,$ in consequence, $\beta+\gamma<\alpha$. 
Now let us prove $\forall \beta,\gamma<\alpha[\beta\cdot\gamma<\alpha]$. We have, using this, that $\deg(\beta\cdot\gamma)\leq\max(\deg(\beta)+\deg(\gamma),\deg(\gamma)+\deg(\beta))<\alpha$, since $\deg(\beta),\deg(\gamma)<\alpha$ and what it was shown above.
Now let us see $\forall \beta,\gamma<\alpha[\beta^\gamma<\alpha].$ Let $\beta'=\deg(\beta)$. Then as $\alpha=\omega^\alpha$, $\beta<\alpha$, but clearly $\alpha$ is limit, so that $\beta'+1<\alpha$. But $\beta^\gamma\leq (\omega^{\beta'+1})^\gamma=\omega^{(\beta'+1)\cdot\gamma}<\omega^\alpha=\alpha$; since $(\beta'+1)\cdot\gamma<\alpha$ by what it was shown above and the uniqueness of Cantor's normal form. 
$(\Leftarrow)$ As $\alpha>\omega$, in particular $\omega^\gamma<\alpha$ for all $\gamma<\alpha$, hence $\omega^\alpha=\lim_{\gamma\to\alpha}\omega^\gamma\leq\alpha,$ but clearly $\omega^\alpha\geq\alpha$.
