Find the integers $n \geqslant 1$ for which $n$ divides $2^{n}-1$.
First, $n=1$ is a solution. We now look for a potential solution $n \geqslant 2$.

Let $p$ be the smallest prime factor of such an integer $n$. Since $2^{n} \equiv 1(\bmod p)$, we know that $p$ is odd and that $\omega_{p}(2)$ divides $n$. However, $\omega_{p}(2)$ also divides $p-1$. But, by the minimality of $p$, the integers $p-1$ and $n$ are coprime. This means that $\omega_{p}(2)=1$, that is, $p$ divides $2-1=1$. Faced with such an absurdity, we conclude that no integer $n \geqslant 2$ is a solution, and therefore that $n=1$ is the unique solution to the problem.