Let x = pi/(2k+1), for k>0.
cosxcos2xcos3x...coskx = (1/2)^k
I've confirmed this numerically for n from 1 to 30. I'm finding it surprisingly difficult using standard trig formula manipulation. Even for the case k = 2, I needed to actually work out cosx by other methods to get the result.
Please let me know if you have a neat proof.