Epsilon numbers

Question

Let $\alpha$ be an ordinal number and define $f_\alpha$ as:

• $f_\alpha(0) = \alpha + 1$
• $f_\alpha(n+1) = \omega^{f_a(n)}$

Let $S(\alpha) = \sup\{f_a(n)\ |\ n \in \omega\}$

Then $S(\alpha)$ is an epsilon number and is the least epsilon number greater than $\alpha$.

Since none of natural numbers are epsilon number, I think $S(n)=S(m)$ for every natural numbers $n,m$. I know that I'm wrong but I don't know why. Please, help.

And I have problem with showing that $m<n\implies S(m)<S(n)$

Answer 1

Since the question was answered in comments and we don't like leaving questions unanswered, I'm adding azarel's comment as a CW answer:

Actually, $S(n)=S(m)$ for all natural numbers $n,m$. Since $\epsilon_0$ is the minimal epsilon number above of them.