We prove the conjecture of Abbott and Katchalski that for every m ≥ 2 there is a positive constant λm such that S(Kd mn) ≥ λmnd−1 S(Kd−1 m ) where S(Kd m) is the length of the longest snake (cycle without chords) in the cartesian product Kd m of d copies of the complete graph Km. As a corollary, we conclude that for any finite set P of primes there is a constant c = c(P) > 0 such that S(Kd n) ≥ cnd−1