If $a\geq 2$, $a\nmid b$, and $a^n-1\mid b^n-1$ for all $n\in\mathbb{N}$, then $b=1$

Question

Let $a,b\in\mathbb{N}$ be such that $a\geq 2$, $a\nmid b$, and $a^n-1\mid b^n-1$ for all $n\in\mathbb{N}$, then $b=1$.

PS: In fact, if we do not assume that $a\nmid b$, then the statement should be $b=a^k$ for some $k\in\mathbb{N}\cup\{0\}$. The above statement can deduce the fact.

Answer 1

This is a solution here which was found in some paper in AMM. it's a very nice proof. but it's very constructive. For example, the construction of $Q_k(x)$, $p_k$ and $r_{k,n}$, it's not so easy to construct them. Maybe there is some direct method.