AHH

\(7^n\div\frac{1000!}{(500!)^2}=N \qquad N\in Z\\ \)

500! = 7*14* ....(71*7) * k                    where k is some integer that is not divisale by 7  (each k is a different interger)

500! = 7(1*2*3*.....*71)k = 7*71! * k

(500!)^2 = 49*(71)!^2 *k

1000! = 7*14* ......  142*7 *k    

\(\frac{1000!}{(500!)^2}\\ =\frac{7(142!)}{7*71!*7*71!}k\\ =\frac{(142!)}{71!*7*71!}k\\ =\frac{(7*14*21*....20*7)}{7(7*14*21* .... 10*7)*(7*14*21* .... 10*7)}k\\ =\frac{(7*20!)}{7(7*10!)*(7*10!)}k\\ =\frac{(20!)}{7*7*(10!)*(10!)}k\\ =\frac{k}{(10!)*(10!)}\\ =\frac{k}{49}\\\)

So I think the biggest  n must be less than or equal to 2          

I could easily have made stupid mistakes though.

Actually I think my answer could easily be rubbish.

Not sure if I understand your question!

1 - if 7^n mod (1000! / 500!^2) =0, then there is no integer solution for n.

2 - if (1000! / 500!^2) mod 7^n =0, then n has the following values:

n = -25, -24, -23.........all the way to n =0.