Prove the Radical of an Ideal is an Ideal

I am given that $R$ is a commutative ring, $A$ is an ideal of $R$, and $N(A)=\{x\in R\,|\,x^n\in A$ for some $n\}$.

I am studying with a group for our comprehensive exam and this problem has us stuck for two reasons.

FIRST - We decided to assume $n\in\mathbb{Z}^+$ even though this restriction was not given. We decided $n\ne 0$ because then $x^0=1$ and we are not guaranteed unity. We also decided $n\notin\mathbb{Z}^-$ because $x^{-1}$ has no meaning if there are no multiplicative inverses. Is this a valid argument?

SECOND - We want to assume $x,\,y\in N(A)$ which means $x^m,\,y^n\in A$ and use the binomial theorem to expand $(x-y)^n$ which we have already proved is valid in a commutative ring and show that each term is in A so $x-y$ is in $N(A)$. The biggest problem is how to approach the $-y$ if we are not guaranteed unity. Does anyone have any suggestions?

Thank you in advance for any insight.

You seem to have everything else, so here’s one approach to the question of why $x\in N(A)$ implies that $-x\in N(A)$.

Since $(-a)b +ab =0b=0$, we see that $(-a)b=-(ab)$, and so $(-x)^n=x^n$ if $n>0$ is even, and it’s $-x^n$ if $n$ is odd. In either case, if $x^n\in A$, then $(-x)^n\in A$, so that $x\in N(A)\Longrightarrow-x\in N(A)$.