jmc

Prime factorize 1560 to find $1560=2^3\cdot 3\cdot 5\cdot 13$. If $n\le 12$, then $n!$ does not contain a factor of 13. However, $13!$ contains a factor of 13, as well as two factors of 5, five factors of 3, and ten factors of 2. Therefore, the least value of $n$ for which 1560 divides $n!$ is $\boxed{13}$.