If $I \subseteq \sqrt{J}$, then there is an $n \in \mathbb{N}$ such that $I^n \subseteq J$

I am stuck at the following exercise:

Let $R$ be a commutative ring with unity and let $I,J$ be ideals in $R$. Show that if $I \subseteq \sqrt{J}$ and $I$ is finitely generated, then there is an $n \in \mathbb{N}$ such that $I^n \subseteq J$.

I know that since $I$ is finitely generated there are $a_1, \ldots, a_k \in I$ such that $$I = Ra_1+\ldots +Ra_k.$$

If I am not mistaken $I^n = \bigg\{ \sum_{i=1}^m b_1\cdot\ldots\cdot b_n \quad \bigg\vert \quad m \in \mathbb{N} \text{ and } b_j \in I \bigg\}$. But I do not see how I should argue from here that $I^n \subseteq J$. Could you please help me?

From the comments:

If $I = (a_1,\dots,a_m)$ then $I^n = (a_1^{i_1}\dots a_m^{i_m} : i_1 + \dots + i_m = n)$.

There is some $n_0$ such that $a_i^{n_0} \in J$ for $i \in \{1,\dots,m\}$. Hence, you need to argue that for large enough $n \ge n_0$, each of the monomials $a_1^{i_1}\dots a_m^{i_m}$ will be divisible by some $a_i^{n_0}$.

E.g. if $I = (a, b)$ then $I^2 = (a^2, ab, b^2)$ and $I^3 = (a^3, a^2b, ab^2, b^3)$ and $I^4 = (a^4, a^3b, a^2b^2, ab^3, b^4)$, etc. So the generators of $I^3$ are each divisible by $a^2$ or $b^2$, the generators of $I^5$ are divisible by either $a^3$ or $b^3$.