The 1991 USAMO Problem 3 offers a concise proof-by-contradiction that if $n$ is any positive integer, then the sequence $2, 2^2, 2^{2^2}, 2^{2^{2^2}}, \dots$ is eventually constant $\pmod n$. However, I don't fully understand the following step in that proof:
Claim: If the sequence $2, 2^2, 2^{2^2}, 2^{2^{2^2}},\dots$ is not eventually constant $\pmod n$, then the sequence of its exponents $1, 2, 2^2, 2^{2^2},\dots$ is not eventually constant $\pmod k$, where $k$ is the eventual period of $2^0,2^1,2^2,2^3,\dots\pmod n$.
Equivalently: If $1, 2, 2^2, 2^{2^2},\dots$ is eventually constant $\pmod k,$ then $2, 2^2, 2^{2^2}, 2^{2^{2^2}},\dots$ is eventually constant $\pmod n.$
Equivalently: If $2, 2^2, 2^{2^2}, 2^{2^{2^2}},\dots$ is eventually constant $\pmod k,$ then this same sequence is eventually constant $\pmod n.$
I'm sorry if this is something obvious/elementary, but would someone please explain why this is true when $n\gt 1?$ How exactly does the claim then depend on $k$ being the eventual period of $2^0,2^1,2^2,2^3,\dots\pmod n?$
The power towers might distract your attention from what is really happening, which is just this:
Proof. By definition $k$ has the property that $(2^i \bmod n)_i$ eventually has period $k$, which is the same as saying that if $A$ is large enough we will have $2^A \equiv 2^{A+k} \pmod n$.
But that means that if we go far enough in the increasing sequence $a_i$ we eventually reach a point where $a_i$ is large enough to this for happen. And if we go far enough for this to happen and for $(a_i)$ to become constant modulo $k$, we will forever have $a_{i+1}=a_i+jk$ for some $j$, and therefore $$ 2^{a_i} \equiv 2^{a_i+k} \equiv 2^{a_i+2k} \equiv \cdots \equiv 2^{a_i+jk} = 2^{a_{i+1}} \pmod n$$