Factorials

$n! = 7! \times 5! \times 3!$

$= 7! \times 5 \times 4 \times 3 \times 2 \times 3 \times 2$

$= 7! \times (4\times 2) \times (3\times 3) \times (5 \times 2)$

$= 7! \times 8 \times 9 \times 10$

$n!=10! \Rightarrow \boxed {n=10}$

n!=7! 5! 3!
Since there are no factor 11 inside,
So n should be smaller than 11.
Since there are 2 factor 5 inside,
So n should be bigger or equal to 10.
10≤n<11.
N is 10