Let $a,b \in \mathbb{N}-\{0,1\}$ be such that $10 \nmid a$. Then, define
$S_a(b) := \nu_{10}(a^{10^{b+1}} - a^{10^b})$ and $D_a(b) := S_a(b) - S_a(b-1)$.
So, $S_a(b)$ counts the number of common trailing digits between $a^{10^{b+1}}$ and $a^{10^b}$, and $D_a(b)$ is the number of new (rightmost) frozen digits added when $b$ increases by $1$.
Now, I conjecture that, for each integer base $a \geq 2 : 10 \nmid a$, the sequence $S_a(b)$ becomes linear with slope $1$ (eventually) after a small transient.
There exists a constant $C(a)$ such that
$S_a(b) = b + C(a)$, and hence $D_a(b) = 1$, $\forall b \geq 3$ (e.g., $b(2)=b(12)=b(807)=3$).
Question. Is it true that, for every $a>1$ with $10\nmid a$, $D_a(b)=1$ holds for all $b\geq 3$ ?
If yes, can this be proved directly from LTE or multiplicative-order arguments?
Remark. From a previous paper of us involving the so called "constancy of the congruence speed of tetration" (see Def. 2.1 of Number of stable digits of any integer tetration), a stronger non-uniform bound we already proved is that $D_a(b)=1$ for all $b\geq 2+\tilde{\nu}(a)$, where (denoting by $\nu_p(a)$ the $p$-adic valuation of $a$) $$\tilde{\nu}(a)= \begin{cases} \nu_5(a-1), & a\equiv1 \pmod5\\ \nu_5(a^2+1), & a\equiv 2,3\pmod5\\ \nu_5(a+1), & a\equiv 4\pmod5\\ \nu_2(a^2-1)-1, & a\equiv 5\pmod{10} \end{cases}.$$
Thus, my question is whether the uniform threshold $b \geq 3$ always suffices.