lean_workbook_plus_51342

Then $2^{2^n}\equiv -1\pmod{p}\implies 2^{2^{n+1}}\equiv 1\pmod{p}$ , so $\text{ord}_p(2)\mid 2^{n+1}$ , so $\text{ord}_p(2)=2^k$ with $0\le k\le n+1$ . If $k<n+1$ , then $2^{2^n}\equiv 1\pmod{p}$ , but also $2^{2^n}\equiv -1\pmod{p}$ , so $1\equiv -1\pmod{p}$ , so $p=2$ , contradiction, so $k=n+1$ .