Proving \epsilon_0 is the Smallest Epsilon Number

[MathematicalPhysicist]

prove that $$\epsilon_0$$ is an $$\epsilon$$ number and that it's the smallest number.
$$\epsilon_0=\lim_{n<\omega}\phi(n)$$
$$\phi(n)=\omega^{\omega^{\omega^{...^{\omega}}}}$$ where $$\omega$$ appears n times.
an epsilon number is a number which satisfies the equation $$\omega^{\epsilon}=\epsilon$$.
for the first part of proving that it's an epsilon number i used the fact that $$1+\omega=\omega$$, for the second part I am not sure i understand how to prove it:
i mean if we assume there's a number smaller than $$\epsilon_0$$ that satisfy that it's an epsilon number, then $$\omega^{\epsilon^{'}}=\epsilon^{'}<\epsilon_0=\omega^{\epsilon_0}$$
i know that there exists a unique ordinal such that $$\epsilon_0=\epsilon^{'}+\alpha$$
if $$\alpha$$ is a finite ordinal then $$\epsilon_0=\epsilon^{'}$$ and it's a contradiction, but how to prove it when alpha isn't a finite ordinal?

 

[matt grime]

I don't know if this is at all valid in the theory of ordinals, but surely lims and exponentials commute, hence e_0 is an epsilon number.

Secondly, if e is an epsilon number then e=w^e (=>w) = w^w^e (=>w^w) =... hence e must be greater than w, w^w, w^w, w^w^w,.. and therefore e must be greater thanor equal to the smallest ordinal larger than all of w, w^w, w^w^w, etc which is precisely e_0.

(This is exactly the same as showing that if t is larger than 0.9, 0.99, 0.999,... then t=>1.)