Tetration and p-adic

[Ember Edison]

But the essential result you showed is that:

\[
\lim_{N\to\infty}\,\, ^N a = \overline{a_1a_2\cdots a_N}A\\
\]

Where we repeat to the left; and that means the value is a rational number in p-adic circles; the same way: \(0.\overline{9} = 1\) is a rational number. I'm not the best versed in this shit though; I just know enough to get by if someone starts talking about it . Definitely continue your research though! And you're only doing favours for yourself if you phrase it in p-adic terms

You'd definitely get some eyes on your search for tenure if you proved hands down \(^\infty a \in \mathbb{Q}_p\) for all \(a \in \mathbb{N}\) and \(p\) prime--or even \(a \in \mathbb{Q}\). That's a solid, sexy, result that tenure boards love

This is indeed a result that is very close to the final answer!
The difference between the different "versions" of p-adic is not significant in terms of series representation \( b = \sum\limits_{r \in S} a_r p^r  \).

for example: 
\( \mathbb{Z}_p \to 0\leq r<p \)
\( \mathbb{Q}_p \to S \) is a bounded-below subset of the \( \mathbb{N} \)

...

\( \bar{\mathbb{Q}}_p, \mathbb{C}_p  \to S \) is a more general well-ordered subset of \( \mathbb{Q} \)
\( \Omega_p \to S \) is any well-ordered subset of \( \mathbb{Q} \) with no other restrictions! 

If we can compute the required infinite limit for any \(a_r\), the problem of evaluating \( ^\infty b \) (or my symbol \( sexp(b,\infty) \) ) is solved.