By definition, $α$ is an epsilon number iff $α=ω^α$. Indeed, the $ε$, $ζ$, and $η$ numbers are all epsilon numbers. But is $ω_1$ also an epsilon number? It seems that $ω^{ω_n}=ω_n$ for $n > 0$, which would mean that even ordinals $ω_1$ to $α↦ω_α$ are all epsilon numbers.
Edit: What I meant was ordinals greater than all the $ω_0$ numbers.
Edit 2: Yes, every initial ordinal greater than $ω$.
Thank you.
Well, as written the answer is obviously no: $\omega+1$ is an ordinal greater than $\omega$ which is not an $\epsilon$-number.
However, I suspect you meant to ask whether every cardinal greater than $\omega$ (or if you prefer, every initial ordinal greater than $\omega$) is an $\epsilon$-number. The answer to this, as you suspect, is yes. The key point is that as long as $\alpha$ is infinite we have $\vert\alpha\vert=\vert\omega^\alpha\vert$ (where ordinal exponentiation is being used here, of course). This means that any uncountable cardinal is a limit of $\epsilon$-numbers, and any limit of $\epsilon$-numbers is again an $\epsilon$-number.