 Solution

Fix $n\ge 1$ 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 $M=(u_1,\dots ,u_r)$. It is nonzero: if $0\ne a\in I$, then $a^n\ne 0$ because $R$ is an integral domain.

The equality $M=IM$ allows us to writeLet $A=(a_{ij})$ and $B={\mathrm{Id}}_r-A$. Then $B(u_1,\dots ,u_r{)}^{\mathsf{T}}=0$. The adjugate matrix identity $\mathrm{adj}(B)B=(\det B){\mathrm{Id}}_r$ implies that $\det B$ annihilates every $u_i$. In the determinant expansion of ${\mathrm{Id}}_r-A$, the constant term is $1$ and every other term contains an entry of $A$, soThis is the determinant trick. Since some $u_i\ne 0$, 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: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 $x\in \sqrt{I^n}$. Thusand $J=I,I^2,I^3,\dots$ 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.