I will prove this by using induction. You can manually check that the statement (\(5|n^5-n\)) is true when \(n=1\), \(n=2\) and \(n=3\). Now let \(n = k + 1\), and let's expand \((k+1)^5 - (k+1)\):
$$ (k+1)^5 - (k+1) $$$$ (k^5 + 5k^4 + 10k^3 + 10 k^2 + 5k + 1) - (k + 1) $$
Simplify:
$$ (k^5 - k) + 5k^4 + 10k^3 + 10k^2 + 5k $$
If \(n = 4\), then \(k=3\), and since all the terms above are divisible by 5 when \(k=3\), then \(5|(k+1)^5 - (k+1)\), or in other words \(5|(4)^5-4\). Now if \(n=5\), then \(k = 4\), and since all the terms above are divisible by 5 when \(k=4\), then \(5|(4+1)^5-(4+1)\) or \(5|5^5-5\). We can keep doing this for every number.