How to prove $n^5 - n$ is divisible by $30$ without reduction How can I prove that prove $n^5 - n$ is divisible by $30$?
I took $n^5 - n$ and got $n(n-1)(n+1)(n^2+1)$
Now, $n(n-1)(n+1)$ is divisible by $6$.
Next I need to show that $n(n-1)(n+1)(n^2+1)$ is divisible by $5$.
My guess is using Fermat's little theorem but I don't know how.

$n^5 -n = n^5 + 6n -n = n(n^(4) - 1) + 6n
=n(n+1)(n^2 + 1)(n-1) + 6n
6(n^2 +1) m + 6n
6[{n^2+1+n}]$
by fermat 's theorem** $n^5$  is congruent to  $n\mod 5$
$5|n^5 -n$
hence,
$30|n^5-n$
