(I'll leave this proof to someone else)
So, let's consider the following, seeking contradiction
Quite honestly, I'm not sure if this is how you attack this. Fermat's little Thm. applied to sheds light on the fact that
Here we observe that this bit is divisible by 2 (2 consecutive numbers multiplied together), divisible by 3 (one of three consecutive terms certainly is divisible by 3), divisible by 5 (by virtue of the 0(mod 5)), and divisible by 6 since that's 2*3.
Just offering my $0.02. I'd be interested in a elegant solution to this.