jmc

We find the prime factorization of $80325$, which is $3^3\cdot 5^2\cdot 7\cdot 17$. The largest prime in the factorization is $17$, so $n$ is at least 17. Since there are three factors of $3$, two factors of $5$, and one factor of $7$ in the prime factorization of $17!$, the minimum value of $n$ is $\boxed{17}$.