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For a prime $p$ and $n\ge 1$, the unique residue $y$ modulo $p^n$ satisfying $y\equiv x(\mathrm{mod}p)$ and $y^p\equiv y(\mathrm{mod}{p}^n)$ is the displayed residue. Existence follows from Fermat's little theorem and prime-power contraction under pth powers. Uniqueness follows by raising any other such residue repeatedly to the $p$th power and applying the same contraction to its congruence with $x$. This is the finite-modulus version of a Teichmuller representative, including the zero class.

The prime-power contraction under pth powers is the key step. If $a=b+p^{k}h$ with $k\ge 1$, the binomial theorem givesThe first term is divisible by $p^{k+1}$. Each later term is also divisible by that power because $kj\ge k+1$ for $j\ge 2$. This includes $p=2$ and $k=1$. ThereforeStart with $x\equiv y(\mathrm{mod}{p}^n)$ and apply this implication $r$ times, increasing both the exponent and modulus at each step:The case $r=0$ is precisely the original congruence.

By Fermat's little theorem, $x^p\equiv x(\mathrm{mod}p)$. Apply the same result with initial modulus $p$ and $r=n-1$: