= Prime-power contraction under pth powers
{title2=$a\equiv b\pmod{p^k}\Rightarrow a^{p^r}\equiv b^{p^r}\pmod{p^{k+r}}$}
For a prime $p$ and $k\geq1$, write $a=b+p^kh$. 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 $r\geq0$, including $p=2$. Applied to <Fermat's little theorem>, it yields $a^{p^n}\equiv a^{p^{n-1}}\pmod{p^n}$.
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