119
Largest positive integer n for which n! can be expressed as the product of n-a consecutive positive integers is (a+1)! - 1.
Proof - let the largest of the n-a consecutive positive integers be k. Clearly k cannot be less than or equal to n, else the product of n-a consecutive positive integers will be less than n!
Now, observe that for n to be maximum the smallest number (or starting number) of the n-a consecutive positive integers must be minimum, implying that k needs to be minimum. But the least k > n is n+1.
So the n-a consecutive positive integers are a+2, a+3, … n+1
So we have
(n+1)!/(a+1)! = n! => n+1 = (a+1)! => n = (a+1)! -1
for a=4 we have n= 5!-1=119
