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?

As suggested in the comment by Akiva Weinberger, take $a=5$ and $b=1$.
Then $(k+1)^a-(k+1)^b=(k+1)^5-(k+1)=(k+1)[(k+1)^4-1]$
$=(k+1)(k^4+4k^3+6k^2+4k)=(k+1)k(k^3+4k^2+6k+4)$.
This is divisible by $2$ because  $(k+1)k$ is,
and modulo $5$ this is the same as $(k+1)k(k-1)(k^2+1)$, so it is divisible by $5$ too,
and therefore it is divisible by $10$.
