A large power sum modulo seven

First, $2222\equiv3\pmod{7}$ , and $5555\equiv4\pmod{7}$ . So it becomes $3^{5555}+4^{2222}$ in modulo 7. From Fermat's, we have $3^6$ and $5^6$ are both 1 (mod 7), so it reduces to $3^5+4^2\pmod{7}\rightarrow{259\pmod{7}\equiv0\pmod{7}}$ .

Source: InternLM Lean-Workbook, record lean_workbook_45641; Apache-2.0. Complete source proposition preserved.