Ok, problem solved. We just needed to resume the original tetration base by multiplying those numbers by \(20\), adding \(5\) to the result so that we find the corresponding solution from \(y^5=y\), which is the \(10\)-adic integer \( \alpha_{25}=\{5^{2^n}\}_{\infty}=\dots 92256259918212890625 \).
For every radix \(r\geq 2\), the congruence speed of tetration stabilizes at a positive constant for all bases \(a>1\) with \(\mathrm{rad}(r )\nmid a\) if and only if \(r\) is squarefree. (NNTDM 32(3), Appendix B)
