Problem - 4182
Let $n$ be an odd integer greater than $1$, then $n$ is the multiplicative order of $2$ modulo $(2^n-1)$.
It plainly holds that $2^n\equiv 1\pmod{2^n-1}$. Therefore, it is sufficient to show that for any positive integer $k$ less than $n$, $2^k\not\equiv 1\pmod{2^n-1}$ which is equivalent to $(2^n - 1) \not\mid (2^k-1)$. This obviously holds because $(2^n -1 ) > (2^k -1) > 0$.