lean_workbook_plus_2164

For all prime $p \neq 2$ , there exists some $n \in \mathbb N$ for which $p | 2^n-1$ . This is true. Since $\gcd(2, p) = 1$ , we have that $2^{\varphi(p)} = 1\pmod p$ . Then let $n = p - 1$ so $2^{p-1} = 1 \pmod p$ , thus $p | 2^{p-1} - 1$ .