$I,J$ ideals in $k[x_1,...,x_n]$, $J\subseteq \sqrt{I}$, then $J^r \subseteq I$ for some $r>0$.
I found this as an exercise in the book Ideals, Varieties and Algorithms (Cox, Little, O'Shea). It has the tip: use the Hilbert Basis Theorem. So I chose a system of generators for $\sqrt{I}$, $\{f_l\}_{l\in L}$ such that $\{f_h\}_{h\in H\subset L}$ was a base for $J$. So then I have to show that some $r$ exists so that all the products of $r$ elements of the system of generators for $J$ are in $I$. Each $f_l$ belongs to $\sqrt{I}$, thus, for some $N$, $f_l^N \in I$ (the system of generators is finite, so we can choose the biggest $N$). But I am not sure about the products of different $f_l$'s. I was thinking about considering $(\sum f_l)^N$ for different amounts of $f_l$, but I am not sure that is going to be a way to solve this, and I am not sure if taking $r=N$ is the right idea.
If we actually take $r=NL$ ($L=$ number of elements in a system of generators of $J$), then $\prod_{l=1}^r f_l= \prod_{i=1}^Lf_i^{k_i}$ (in the first product, some factors might be repeated), where some $k_i \geq N$ because $\sum_{i=1}^Lk_i=NL$. Thus, $f_i^{k_i} \in I$ because of the definition of $N$, so the whole product must be in $I$. So we have that every product of $r=NL$ elements of the system of generators of $J$ is in $I$, which means that $J^r \subseteq I$.