Find all natural numbers $n$ such that $n \mid 2^{n}-1$.
We see that $n$ must be odd. Since $n=1$ is a solution, we now assume $n>1$. Let $p$ be its smallest prime divisor, which is therefore odd, and $\omega$ the order of 2 modulo $p$. Then, according to the statement, $\omega \mid n$, and by Fermat's little theorem, $\omega \mid p-1$. However, $p-1$ has only prime divisors strictly smaller than $p$, and thus does not divide $n$ by the minimality of $p$. Therefore, $n \wedge (p-1)=1$ and thus $\omega=1$. Hence, $2 \equiv 1 \pmod{p}$, which is absurd. Therefore, $n=1$ is the only natural number such that $n \mid 2^n - 1$.