Prove

[Sudip Deb new]

Prove that (n^5 - n) is divided by 5 .

[HandaramTheGreat]

$n^5-n=n(n^4-1)=n(n^2+1)(n^2-1)$
last digit of a square can be only 1, 4, 9, 6, 5 or 0...
if it's 1, 4, 9, or 6, then $(n^2+1)$ or $(n^2-1)$ is obviously divisible by 5...
if it's 5 or 0, then $n$ should be divisible by $5$...

if you put 2 dollar sign in both side of (n^5-n), it would be like this $(n^5-n)$...

[Moon]

Also you can always use Fermat's Little Theorem, which states that for all $n\in\mathbb{N}$ and any prime $p$ \[ n^p \equiv n \pmod{p} \]
When $gcd(n,p)=1$, you can divide the equivalence by $n$ and get $n^{p-1} \equiv 1 \pmod{p}$.