Find all $n \in \mathbb{N}^{*}$ such that $n$ divides $2^{n}-1$.
Let $n>1$ such that $n$ divides $2^{n}-1$, and $p$ be the smallest prime divisor of $n$. We denote $d$ as the order of 2 modulo $p$: we know that $d$ divides $p-1$, and on the other hand, from the statement $p \mid 2^{n}-1$ so $d$ divides $n$, thus $d$ divides $\text{GCD}(n, p-1)$.

However, since $p$ is minimal, $p-1$ is smaller than all prime divisors of $n$, so $n$ and $p-1$ have no common prime divisors, thus their $\text{GCD}$ is 1, so $d=1$. In other words, $2 \equiv 1(\bmod p)$, so $p=1$, which is absurd, so the only solution is $n=1$.