Prove that the last digit of $4n^5-5n^2+n$ is $0$ 
Prove that the last digit of $4n^5-5n^2+n$ is $0$ for all natural $n$

My attempt:
$$4n^5-5n^2+n\overset{?}\equiv 0\pmod {10}$$
using Fermat's little theorem 
$$4n\cdot\pmod 5 -5n\pmod 2+n\overset ?\equiv 0 \pmod{10}$$
I am suck here

You want to prove that $4n^5-5n^2+n\equiv 0\pmod {10}$.
So prove that $4n^5-5n^2+n\equiv 0\pmod {2}$, which is equivalent to $-n^2+n\equiv 0$, which is always true because $n^2\equiv n\pmod {2}$.
Then prove that $4n^5-5n^2+n\equiv 0\pmod {5}$, which is equivalent to $-n^5+n\equiv 0$, which is always true because $n^5\equiv n\pmod {5}$.
