Showing that $I^nM=0$ for some $n$ when $M$ is a finitely generated module and $Supp_A(M) \subseteq \mathcal{V}(I)$

Question

If $M$ is a finitely generated $A$-module, where $A$ is noetherian, and if $I$ is an ideal of $A$ such that $Supp_A(M) \subseteq \mathcal{V}(I)$, I want to show that $I^nM = 0$ for some $n$. An obvious approach is to use the fact that $A$ is noetherian to show that there is an $n$ such that $I^n = I^{n+k}$ for all $k$, since $I \subseteq I^2 \subseteq I^3 \subseteq \ldots$ is an ascending chain of ideals. Also, since $M$ is finitely generated, we have $Supp_A(M) = \mathcal{V}(Ann_A(M))$, we have the fact that every prime ideal which contains the annihilator of $M$ also contains $I$. I'm not quite sure how to use all this information to show that $I^n$ annihilates $M$. Now that I type it, it doesn't seem true, since $I^n$ would have to be contained in $Ann_A(M)$. Any hints are appreciated.