The last digit of $n^5-n$ What will be the last  digit of
$$n^5 - n \bmod 1000,$$
where $n$ is a natural number and $m \bmod 1000$ is the remainder when $m$ is divided by $1000$.

The unit digit of $n^{4k+1}$ is same as the unit digit of $n$
This is because the unit digit is periodic with period $4$,or $2$, or $1$, For instance
$2^1=2 _{U.D=2}$
$2^2=4_{U.D=4}$
$2^3=8_{U.D=8}$
$2^4=16_{U.D=6}$
$2^5=32_{U.D=2}$
$2^6=64_{U.D=4}$
Hence we observe that the unit digit is $2,4,8,6$ and it oscillates with period $4$, in case of $2^n$,
For numbers like $5^n$,$11^n$,$6^n$ this period is $1$,
and for $9^n$ the period is $2$.
Hence the last digit of $n^5-n$ is $0$.
