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.

According to link#1, the product of 5 consecutive integers divisible by 5! and the product of 3 consecutive integers divisible by 3!.
Now $n(n-1)(n+1)(n-2)(n+2) = n^5-5n^3+4n $
$n^5-n= n^5-5n^3+4n + 5(n^3-n)$
=$n(n-1)(n+1)(n-2)(n+2)  -5n(n-1)(n+1)$
So, the first part is divisible by 5! and the 2nd part is by 5(3!)
