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For a prime $p$ and $k\ge 1$, write $a=b+p^{k}h$. The binomial theorem shows $a^p-b^p$ is divisible by $p^{k+1}$: the linear term contains the extra factor $p$ and every higher term contains at least $p^{2k}$. Iteration gives the displayed modular congruence for every $r\ge 0$, including $p=2$. Applied to Fermat's little theorem, it yields $a^{p^n}\equiv a^{p^{n-1}}(\mathrm{mod}{p}^n)$.