= Solution
{c}
Fix $n\geq1$ and suppose, for a contradiction, that $I^n=I^{n+1}$. Set $M=I^n$. Since $R$ is a <Noetherian ring>, this <ideal> is a <finitely generated module>, say $M=(u_1,\ldots,u_r)$. It is nonzero: if $0\ne a\in I$, then $a^n\ne0$ because $R$ is an <integral domain>.
The equality $M=IM$ allows us to write
$$
u_i=\sum_{j=1}^r a_{ij}u_j,\qquad a_{ij}\in I.
$$
Let $A=(a_{ij})$ and $B=\operatorname{Id}_r-A$. Then $B(u_1,\ldots,u_r)^{\mathsf T}=0$. The <adjugate matrix> identity $\operatorname{adj}(B)B=(\det B)\operatorname{Id}_r$ implies that $\det B$ annihilates every $u_i$. In the <determinant> expansion of $\operatorname{Id}_r-A$, the constant term is $1$ and every other term contains an entry of $A$, so
$$
\det B=1-b\quad\text{for some }b\in I.
$$
This is the <determinant trick>. Since some $u_i\ne0$, <multiplicative cancellation in an integral domain> gives $1-b=0$. But then $1=b\in I$, contradicting $I\ne R$. We have proved the <strict descent of powers of a nonzero proper ideal in a Noetherian domain>:
$$
\boxed{I\supsetneq I^2\supsetneq I^3\supsetneq\cdots.}
$$
All these <powers of an ideal> have the same <radical of an ideal>. Indeed, $I^n\subseteq I$ gives $\sqrt{I^n}\subseteq\sqrt I$. Conversely, if $x^m\in I$, then $x^{mn}\in I^n$, so $x\in\sqrt{I^n}$. Thus
$$
\boxed{\sqrt{I^n}=\sqrt I\quad(n\geq1),}
$$
and $J=I,I^2,I^3,\ldots$ gives infinitely many distinct <ideals> with the prescribed <radical of an ideal>. If the convention $0\in\mathbb N$ is used, the additional inequality $I^0=R\ne I$ follows immediately from properness.
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