Commutative Rings and nilpotent

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Here is some background information about the question I'm working on:

If $A$ is a commutative ring, an element $a \in A$ is called nilpotent if $a \neq 0$ but $a^m = 0$ for some positive integer $m$.

My question is the following:

Suppose that $n > 1$ is a positive integer which is not divisible by the square of any prime. Show that $\mathbb{Z_n}$ does not contain any nilpotent elements.

The question actually has two parts, but I solved the first part already.The first part of the question says to find an example of a nilpotent element in the ring $\mathbb{Z_8}$ and I chose $a = 4$ and $m = 2$. In $\mathbb{Z_8}$, $\bar 8 = \bar 0$. How can I extend what I know and apply it to my original question?

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Let $n$ be such that no square of a prime divides it, and consider the quotient map $\mathbb Z\to \mathbb Z_n$. The kernel of the map is $n\mathbb Z$, the multiples of $n$, and so we can rephrase our problem as follows: Show that if $n|r^k$ for some $k$, then $n|r$. This, in turn, can be shown by noting that, since no square of a prime divides $n$, $d$ divides $n$ if and only if $p|d$ for every prime $p$ such that $p|n$.

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This is exercise $26$ here on page $6$, together with a solution. Use $(c)$ and prove the claim.