How do you prove that $ n^5$ is congruent to $ n$ mod 10? How do you prove that $n^5 \equiv n\pmod {10}$ Hint given was- Fermat little theorem. Kindly help me out. This is applicable to all positive integers $n$

By Fermat's Little Theorem, $n^5 \equiv n \pmod 5$
Also $n \equiv 0 \ or \ 1 \pmod 2 \implies n^5 \equiv n \pmod 2$
Chinese Remainder Theorem guarantees the presence of a solution $\pmod {10}$ as $(2,5) = 1$
From the second congruence, $n^5 = 2k + n$
Substitute into the first congruence, $2k+n \equiv n \pmod 5 \implies 2k \equiv 0 \pmod 5 \implies k \equiv 0 \pmod 5$
So $k = 5m$
Substitute that into second congruence, $n^5 = 2(5m) + n  = 10m +n$
Therefore, $n^5 \equiv n \pmod {10}$ $\rm{(QED)}$
