Let $a,b$ be integers not divisible by a prime $p$, show that if $ax^p \equiv b\pmod {p^2}$ is solvable then, $ax^p \equiv b\pmod {p^n}$ is solvable.
What I've tried is by letting $x = w + v$, then $p^2 | ax^p - b \implies p | a(w + v)^p - b$. Now since $p | \binom{p}{k}$ for $k = 1,2,...p-1$, then this implies that $p | a(w^p + v^p) -b$. However now I'm stuck, am I on the right direction? if not any hints on what I should try would be much appreciated, thanks!