you can actually see this in action, and it makes lots more sense:
all 1-cycles are the identity, so we can disregard them (the map that takes 4-->4, and n-->n for all n ≠ 4, is the map that takes n-->n for all n).
so (1 5 6)(2 3)(4) = (1 5 6)(2 3).
[(1 5 6)(2 3)]^2 = (1 6 5) <--- here the 2-cycle factor "hits" the identity ( )
[(1 5 6)(2 3)]^3 = (2 3) <--- here the 3-cycle factor is the identity
[(1 5 6)(2 3)]^4 = (1 5 6) <--- 2-cycle is the identity
[(1 5 6)(2 3)]^5 = (1 6 5)(2 3)
[(1 5 6)(2 3)]^6 = ( ) <--- both factors are the identity.
powers of disjoint cycles are easy to compute, because they (the disjoint cycles) commute with each other,
(1 5 6)(2 3) = (2 3)(1 5 6).