Computing $2016$ using basic operations on the fewest integers, in sequence

Question

Using the operators $$+,-,\div,\times,\exp,(,),!$$ what is the least $n$ to come up with the number $2016$ using the sequence of numbers $1,2,3,\ldots,n$ in that order. You cannot combine numbers, so you cannot do $2~~3=23$ and you cannot negate values.

My solution consists of $10$ numbers. I want to see if someone can come up with the least use of operators in their answer.

Good luck! If no one is able to get less than $10$ numbers, I will post my answer.

Answer 1

Well, this is a 9 number solution.

$1-(2!\times 3! \times 4! \times (5-6) \times 7) +8-9=2016$

I have the feeling that this is not the smallest one, but this is the smallest I can find at the moment.

Answer 2

$4$ number solution: $$\left((1+2)!\right)!+(3!)^4=2016$$

Answer 3

Here is a 7-number solution:

$((1 + 2 * 3) * 4! - 5!) * 6 * 7$