Show that $30$ divides $n^{5} - n$, where n is an integer

Question

I started to think about this problem and then factored $n^5 - n$ to $(n^2 - 1)(n^2 + 1)(n)$, and later to $(n-1)(n)(n+1)(n^2 + 1)$. I know that $(n-1)(n)(n+1)$ is divisible by $6$, but it is not that case $5$ divides $n^2 + 1$ for any integer $n$, so i can´t use the multiplication property. Can anyone help me finish this proof?

Answer 1

This is a particular case of fermats little theorem

Answer 2

Hint:

$$(n^2-1)(n^2+1)n=(n^2-1)(n^2-4+5)n=\underbrace{(n-2)(n-1)n(n+1)(n+2)}_{\text{product of } 5\text{ consecutive integers}}+5n(n^2-1)$$

Answer 3

One of the other answers requires that you know something - the other is very clever. A hint for a proof that doesn't require either non-trivial theorems or cleverness: $n=5k+j$, where $j\in\{0,1,2,3,4\}$.