Every zero-divisors in $Z_{p^{n}}$ is a nilpotent element, where p is a prime.
$\mathbb Z_{p^{n}}$ : set of integers modulo $p^{n}$
nilpotent : $a^{n} = 0,\ \exists n \in \mathbb Z^{+}$
Note that $a \in \mathbb Z_{n}$ is zero-divisor if and only if $gcd(a, n) = d,\ (d > 1)$.
Since that $d \mid p^{n}$, it implies $d = p^{k},\ (k < n)$.
$a = ld = lp^{k}$
$a^{n} = (lp^{k})^{n} = (l)^{n}(p^{k})^{n} = (l)^{n}(p^{n})^{k} = 0$.
nilpotent : $a^{n} = 0,\ \exists n \in \mathbb Z^{+}$
Note that $a \in \mathbb Z_{n}$ is zero-divisor if and only if $gcd(a, n) = d,\ (d > 1)$.
Since that $d \mid p^{n}$, it implies $d = p^{k},\ (k < n)$.
$a = ld = lp^{k}$
$a^{n} = (lp^{k})^{n} = (l)^{n}(p^{k})^{n} = (l)^{n}(p^{n})^{k} = 0$.
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