Nilpotent elements in $\mathbb{Z}_n$
Solution 1:
Saying $n \mid a^k$ implies that $\{$ prime divisors of n $\} \subset \{$ prime divisors of $a^k\} = \{$ prime divisors of a $\}$
Assume now that $n$ has only simple prime divisors, and wonder if it is actually possible that $n \mid a^k$ without $n \mid a$ holding
Solution 2:
Let $p$ be a prime that divides $n$. Then $p\mid a^k$, hence $p\mid a$. If $p_1, \ldots,p_m$ are the prime divisors of $n$, this implies $p_1p_2\cdots p_m\mid a$. If $n$ is squarefree this divisor is $n$ itself.