Divisibility of $10^n-1$ by $7$, $13$, $91$ and $819$

This is Problem 4.108 of N. B. Alfutova and A. V. Ustinov, Algebra and Number Theory (MCCME, 2002), Chapter 4, §4 “Theorems of Fermat and Euler”. The problem asks for an exponent $n$ such that $10^n-1$ is divisible by (a) $7$, (b) $13$, (c) $91$, (d) $819$; the book's answer is $n=6$.

The formal statement records the complete answer. For every natural number $n$ and for each modulus $d\in\{7,\,13,\,91,\,819\}$,

In particular $n=6$ is the smallest positive exponent that works in all four cases (note $91=7\cdot 13$ and $819=9\cdot 7\cdot 13$).

The exercise concerns the multiplicative order of $10$ modulo small primes, which governs the period length of decimal fractions such as $1/7$ and $1/13$.

Formalization Note The four parts are combined into one conjunction of equivalences, quantified over all $n\in\mathbb N$ (the case $n=0$ is harmless: $10^0-1=0$ is divisible by everything and $6\mid 0$). The subtraction $10^n-1$ is natural-number subtraction, which is exact because $10^n\ge 1$.