Mechanical definition of ordinals It seems that one can construct ordinals from bottom up by successively introducing a new symbol each time a limit is taken:
$$1,\ 2,\ \ldots,\ \omega,\ \omega +1,\ \omega +2,\ \ldots,\ \omega\cdot 2,\ \omega\cdot 2 +1,\ \ldots,\ \omega^{2},\ \ldots,\ \omega^{3},\ \ldots\ \omega^{\omega},\ \ldots,\ \omega^{\omega^{\omega}},\ \ldots, \epsilon_{0},\ \ldots$$
Can this be taken as a (mechanical) definition of ordinals? More abstract definitions like "an ordinal is a transitive well-ordered set satisfying certain properties" are much more appealing to me. Is this mechanical definition sufficient to prove things like "each well-ordered set is order isomorphic to exactly one ordinal?"
 A: As remarked in the comments, this is far from sufficient to cover even the countable ordinals. Personally, I see the problem with the three dots at the end which imply both an undefined idea of continuing this sequence, as well something that will terminate after at most $\omega_1$ many steps.
I imagine you might get to some large countable ordinals, perhaps $\epsilon_{\epsilon_0}$ or even higher. However this will terminate long before $\omega_1$.
Why is that a problem? Well, of course that we know about well-ordered sets whose order type is uncountable. But think of this reason: $\mu_0=\{\text{all those ordinals you wrote above}\}$, ordered by $\in$ this would be a transitive and well-ordered set. However it is not isomorphic to any of its members.
More generally, if you have a process which can only be used to generate set many ordinals then this set itself would be a well-ordered set not isomorphic to any of them. This would essentially break the theorem that every well-ordered set is isomorphic to exactly one ordinal.
