(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.

-Andy