Last Digits of a Tetration I was studying tetrations, or "power towers", and I found a decently well-known fact. The last $k-1$ digits of $^k 3 = 3^{3^{\vdots^{3}}} (k \text{ threes)}$ remain constant, for all numbers $^a 3$ with $a \ge k$ (see here for more). Why is this true? The link shows an ad-hoc proof for the last two digits, but how can we tackle larger cases? For example, how can we prove that the last 10 digits of $^T 3$ remain constant for all $T \ge 11$?
 A: Let's start by the first simple thing, for any positive integer $a$ and for any positive integers $q$ we have:
$$\forall k \ \ \ \ \ a\equiv a-kq \mod q $$
taking this to the next level we have:
$$\forall k \ \ \ \ \ a^{a}\equiv a^{a-k\varphi(q)} \mod q $$
and so on :$\displaystyle$
$$\forall k \ \ \ \ \ a^{\displaystyle a^{a}}\equiv a^{\displaystyle a^{a-k\varphi(\varphi(q))}} \mod q $$
Using the notation:
$$\exp_a^n(x) = a^{a^{\cdot^{\cdot^{a^x}}}} \text{with n "a"s}.$$
and recursively we can prove that:

$$\forall k,x \ \ \ \ \  \exp_a^n(x)\equiv \exp_a^n(x-k\varphi^n(q)) \mod q \tag{1}$$

As a consequence to this if $\varphi^m(q)=1$ then:
$$\forall k\geq m\ \ \ \ \ \ \displaystyle ^ka\equiv \displaystyle ^{m-1}a \mod q$$
Now if we return to your question we can notice that $$\varphi^{T_{10}}(10^{10})=1$$ for $T=30$(In general $T_n=3n$) and this gives us your equation for $k\geq 30$ and you can verify that the first elements are equal.
This not very sophisticated, but we can obtain the same reult in $(1)$ if we replace $\varphi$ by the Carmichael function $\lambda$ (the equation remains true for any $a$ , the $m$ does not depend on $a$), but we can furthermore replace $\varphi$ by the order of the element $a$  in the base $q$ denoted by $ord_a(q)$ notice that:
$$\lambda^{l_{10}}(10^{10})=1 \text{ for } l_{10}=12$$
which implies directly your equation and implies that your assertion will always be true if your replace $3$ by any other integer coprime to $10$
A: Integer tetration (for simplicity's sake, let us assume radix-$10$) is characterized by a very special property concerning the difference between the number of (rightmost) frozen digits going from $a \uparrow \uparrow b$ to $a \uparrow \uparrow (b + 1)$ and the number of (rightmost) frozen digits going from $a \uparrow \uparrow (b - 1)$ to $a \uparrow \uparrow b$: for any given base $a : a > 1$ which is not congruent to $0$ modulo $10$, the number of "new" stable digits for any unitary increment of $b$ is constant starting from a sufficiently large value $b(a)$ (i.e., a sufficient but not necessary condition is to take $b > a$ - for details, see DOI: 10.7546/nntdm.2021.27.4.43-61).
Moreover, it is possible to find the exact number of new frozen digits at a given height of any integer tetration by studying the $2$-adic or the $5$-adic valuation of simple manipulations of $a$ (i.e., $v_2(a + 1)$, $v_5(a + 1)$, $v_2(a - 1)$, $v_5(a - 1)$, $v_5(a^2 + 1)$) depending on the congruence modulo $20$ of the tetration base $a$. We could refer to the aforementioned value (which exists and is unique for any $a : a > 1 \wedge a\not\equiv (\mod 10)$) as "constant congruence speed of $a$" or simply $V(a)$.
