Why is $\varepsilon_1$ not smaller?

Question

I'm thinking that $\varepsilon_1$ could in theory be $lim[\omega+1,\omega^{\omega+1},\omega^{\omega^{\omega+1}},...]. $

This, in my opinion, would be smaller than the normal definition of $\varepsilon_1$

Answer 1

If $\alpha$ is properly smaller than $\varepsilon_0$, $\omega^\alpha$ is again properly smaller than $\varepsilon_0$; so this limit can be, at most, $\varepsilon_0$.

Since $\omega^\alpha$ is strictly monotone in $\alpha$, then as Noah says in the comments, every term in the "usual" sequence of powers of $\omega$ is less than or equal to the corresponding term in the sequence you describe; so the limit is greater than or equal to $\varepsilon_0$.

Taking these facts together, the limit of your sequence is again $\varepsilon_0$.

Answer 2

Your definition does indeed correspond to an $\varepsilon$-number. However, it's not $\varepsilon_1$ - it's just a different definition of $\varepsilon_0$. Specifically, note that we have $$\omega<\omega+1<\omega^\omega<\omega^{\omega+1}<\omega^{\omega^\omega}<\omega^{\omega^{\omega+1}}<...$$ That is, the two sequences $\langle\omega,\omega^\omega,\omega^{\omega^\omega},...\rangle$ and $\langle \omega+1,\omega^{\omega+1}, \omega^{\omega^{\omega+1}}, ...\rangle$ "dovetail" and hence have the same limit (namely, $\varepsilon_0$).

This sort of dovetailing rules out a lot of seemingly-plausible ways to describe $\varepsilon_1$ as a "small" ordinal - the twin requirements of $\varepsilon$-number-ness and being strictly bigger than $\varepsilon_0$ wind up making $\varepsilon_1$ quite large indeed.