Prove that $10 | (n^a - n^b)$. $n$ is a positive integer. Prove that there exists positive integers $a$ and $b$, $(a > b)$ such that $10 | (n^a - n^b)$. 
I have tried to prove this by induction on $n$, but I get stuck at the induction step trying to prove it for $n = k + 1$, not knowing how to expand $((k + 1)^a - (k + 1)^b)$. Is this the wrong approach to solving this problem, or am I missing something here?

We have $n^{a} - n^{b} \equiv 0\pmod 2$ since if $n \equiv 1 \pmod 2$ , then $ n^{a} - n^{b}\equiv 1-1 \equiv 0\pmod 2$ and if $n\equiv 0 \pmod 2, n^{a} - n^{b}\equiv0-0 \equiv 0\pmod 2$.
If $5|n$, then $n^{a}-n^{b}\equiv0-0\equiv 0 \pmod 5$ and so $10|n^{a}-n^{b}$.
If $5 \nmid n$, then $n^{4} \equiv 1 \pmod 5$ by Fermat's little theorem so that $n^{8} \equiv 1 \pmod 5 \implies$
$n^{8}-n^{4} \equiv 0 \pmod 5$ and altogether, in this case, $10|n^{a}-n^{b}$ as well where $a=8$ and $b=4$.
