A number theory problem by Julian Yu

$2016=2^5\times 3^2\times 7$
To meet the requirement for $2$ , $n \ge 8$ , since $2, 4, 6$ multiplied together only gives a power $4$ .
Similarly, requirement for $3$ is $n \ge 6$ and that for $7$ is $n \ge 7$
Taking the strictest, $n \ge 8$

Note that $2016 = 288 \times 7$

As $6! = 720$ and 288 cannot divide 720, we need two more multipliers to $6!$ , which is $7\times 8$ , making the value $8! = 40320$ , and, when divided by 2016, gives $20$ as an answer...