Let's  consider having c cycles of size k, d > c cycles of size k+1, and some number of largest cycles of size p >k+1. Let's shift an element out of d-c of the k+1 cycles, and put them on one p cycle. Just considering cycle lengths, in the product this represents a differential of  1 + (d-c)/p against  (1+1/k)^(d-c). If we started with b many p cycles, we now have b-1, and so the product is reduced by an additional factor of b. So "shifting the excess" to a large enough cycle results in a reduction in denominator. Thus almost any cycle that optimizes the product will have more smaller cycles than larger cycles, with all lengths between 1 and l represented, followed by a solitary cycle of length p or more possibly larger than l+1.

(In case we have p=k+1, a similar argument with d-c>1 also works, as does the case when there is more than one cycle of maximal length. So, with the exception of a gap between l and p, we have cycles represented in decreasing number as length grows.)

So with care, the above can be turned into a proof of Aaron's unsurprising result.  Now let us see if we can predict how  big is l, the length of the  largest (but one) cycle .

Suppose the two largest cycle lengths are l and p,  with l less than p, and assume they occur uniquely. We split the l cycle into a 1 cycle and increment the p cycle by l-1. If we started with d many 1 cycles, we win (by shrinking the denominator) if (d+1)(p+l-1) is less than or equal to pl. If we have c many cycles of length l, replace pl by cpl in the previous inequality. So we win for sure if (d+1) is less than or equal to l/2.

If we are giving too much attention to one cycles, we can consider having c many k cycles and increasing their count by 1.  The inequality now becomes a win if (c+1)k(p+l-k) is at most lp. In particular, if kc is at most l/2, consider shortening the l cycle to k and adding the excess up.

By considering a number of moves of this type, the search space for a cycle that optimizes the associated product should be readily obtained, even by hand, for large enough n.

Gerhard "Now Show Me Your Moves" Paseman, 2019.06.10.