Prove ${^{b}a} \equiv {^{b+1}} a \pmod {10^{\lfloor{\log_{10} (^{b}a) }\rfloor + 1}} \Rightarrow a=5$ as $a$ and $b$ are two integers greater than $1$

Question

$\DeclareMathOperator\len{len}$Let $a, b \in \mathbb{N} -\{ 0, 1 \}$ and define ${^{b}a}$ to be $a^a$ if $b = 2$ and $a^{\left(^{b-1}a \right)}$ if $b \geq 3$ (e.g., ${^{3}5} = 5^{\left( 5^5 \right)} = 5^{3125})$.
Then, let $\len({^{b}a}) := \lfloor{\log_{10} (^{b}a) }\rfloor + 1$ indicate the number of digits of ${^{b}a}$ (e.g., given $a=5$ and $b=2$, it follows that $\len({^{2}5}) = \len(3125) = 4)$.

Prove that ${^{b}a} \equiv {^{b+1}a \pmod {10^{\len({^{b}a})}}}$ if and only if $a=5$ ($a=5$ is a solution since ${^{2}5} \equiv {^{3}5} \pmod {10^4})$.

(The present question is related to this MSE thread, but here we are making a stronger assumption on the stated congruence, removing at the same time the previous $b=2$ condition, as $b$ is now free to run over the integers greater than $1$.)

Answer 1

Extending your definition of ${^{b}}a$ to have it be just $a$ for $b = 1$, so ${^{b}}a$ is $a^{({^{b-1}}a)}$ for $b \ge 2$, then your equivalence relation becomes

$$a^{({^{b-1}}a)} \equiv a^{({^{b}}a)} \pmod{10^{\operatorname{len}({^{b}a})}} \;\to\; a^{({^{b-1}}a)}\left(a^{({^{b}}a - {^{b-1}}a)} - 1\right) \equiv 0 \pmod{10^{\operatorname{len}({^{b}a})}} \tag{1}\label{eq1A}$$

In particular, consider

$$a^{({^{b}}a - {^{b-1}}a)} - 1 \equiv 0 \pmod{2^{\operatorname{len}({^{b}a})}}, \;\; a^{({^{b}}a - {^{b-1}}a)} - 1 \equiv 0 \pmod{5^{\operatorname{len}({^{b}a})}}\tag{2}\label{eq2A}$$

As you mention in your comment, it's relatively trivial to show that $10 \nmid a$, so have that be an added restriction. In addition, since $a = 5$ is a solution, also have $a \neq 5$. Next, since $\operatorname{len}(a^a) \ge a - 1$ for all $a \ge 2$, then my answer in the Math SE you linked to proves $a = 5$ is the only solution for $b = 2$. As also shown in that other answer, both equations in \eqref{eq2A} don't hold for any of those other values of $a$, but at least one equation must hold for \eqref{eq1A} to be true.

Assume neither equation in \eqref{eq2A} is true for any $a$ under consideration for $b = k$ for some $k \ge 2$, so \eqref{eq1A} is also not true. With $b = k + 1$, consider one or both equations in \eqref{eq2A} depending on if $2$ or $5$ is a factor of $a$. Using the Lifting-the-exponent lemma (LTE lemma), with the first one if $2 \nmid a$, since $p = 2$ and the exponent is even, we get

$$\nu_2(a^{({^{k+1}}a - {^{k}}a)} - 1) = \nu_2(a - 1) + \nu_2(a + 1) + \nu_2({^{k+1}}a - {^{k}}a) - 1 \tag{3}\label{eq3A}$$

Using that $\nu_2(a - 1)$ or $\nu_2(a + 1)$ is $1$, let $c$ be the value of $a - 1$ or $a + 1$ where it's $\nu_2$ value is $\gt 1$. Also, from the induction hypothesis in \eqref{eq2A}, we have $\nu_2({^{k+1}}a - {^{k}}a) \lt \operatorname{len}({^{k}a})$. Thus, the RHS of \eqref{eq3A} then becomes

$$\nu_2(c) + \nu_2({^{k+1}}a - {^{k}}a) \lt \nu_2(c) + \operatorname{len}({^{k}a}) \lt \operatorname{len}({^{k + 1}a}) \tag{4}\label{eq4A}$$

If $5 \nmid a$, then with the second equation in \eqref{eq2A}, we need to consider $a \mod{5}$. If that modular value is $1$, we can just use $a$, else with $2$ or $3$, we need to use $a^4$, while if it's $4$, then $a^2$ is required. Let $j$ be the minimum required exponent. If $j \nmid {^{k+1}}a - {^{k}}a$, then the second equation in \eqref{eq2A} doesn't hold. Otherwise, let ${^{k+1}}a - {^{k}}a = jm$, so we then have

$$\nu_5\left((a^{j})^m - 1\right) = \nu_5(a^j - 1) + \nu_5(m) \tag{5}\label{eq5A}$$

Similar to \eqref{eq4A}, we get

$$\nu_5(a^j - 1) + \nu_5(m) \lt \nu_5(a^j - 1) + \operatorname{len}({^{k}a}) \lt \operatorname{len}({^{k + 1}a}) \tag{6}\label{eq6A}$$

This shows that both equations in \eqref{eq2A} don't hold for $b = k + 1$ and, thus, \eqref{eq1A} also doesn't hold. As such, this proves by induction that the only solution to your equivalence relation is as you stated, i.e., $a = 5$.