Given an ordinal $\alpha$, define $\alpha !$ as it follows: $$ \alpha! := \begin{cases} 0! = 1 \\ (\alpha + 1)! = \alpha ! \cdot (\alpha + 1) \\ \lambda! = \left(\sup_{\gamma < \lambda} \gamma !\right) \cdot \lambda &\text{$\lambda$ limit} \end{cases} $$
I was trying to evaluate $\varepsilon_0 !$, where $\varepsilon_0$ is the smallest $\varepsilon$-number (the least fixed point of the function $x \mapsto \omega^x$). I can apply the definition as it follows:
$$ \varepsilon_0 ! = \left(\sup_{\alpha < \varepsilon_0} \alpha ! \right) \cdot \varepsilon_0 $$
now all that remains to be done is to evaluate $\sup\{\alpha ! : \alpha < \varepsilon_0\}$. I claim this to be $\varepsilon_0$, because it's obvious that $\varepsilon_0 = \sup\{\alpha : \alpha < \varepsilon_0\}\leq \sup\{\alpha ! : \alpha < \varepsilon_0\}$, since the map $\alpha \mapsto \alpha !$ is stractly increasing (if I'm not mistaken it's a simple induction). Now I'd like to prove the opposite inequality, which means I'd like to prove that for every $\alpha !$, $\alpha < \varepsilon_0$, there exists an ordinal $\beta < \varepsilon_0$ such that $\beta \geq \alpha!$, in order to do this, I was trying to prove that $\alpha < \varepsilon_0 \implies \alpha ! < \varepsilon_0$, and, since $\varepsilon_0$ is a limit ordinal i can use $\beta = \alpha ! + 1 < \varepsilon_0$. But I'm having some troubles, anyone has a suggestion?