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 cl/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.  If you lose, it won't be by much.

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.

**Update 2019.06.11:**

Here is an attempt at justifying an approximation to the conjecture made in the post.  While not strictly following the conjecture, we show that a good attempt at an optimum starts with n large enough, about m/2 fixed points, about m/4 cycles of length 2, and so on.  I do not claim the result is an absolute optimum. I do suggest that the optimum is not far from this choice for n about 2m. For larger n there is more room to play.

I follow the original post, assume n is at least 2m, and look at the permutation of m cycles with one cycle of length n-m+1. The denominator of (n-m+1)((m-1!)) is easily calculated; what changes in cycle structure can make it smaller?

To start, we borrow from the large cycle to make at least (m-1)/2 cycles of length 2, giving possibly more cycles of length 2 than of length 1. (Don't worry, we will recycle some of these 2-cycles. Hah, I'm so funny.) Indeed, when we borrow about (m-1)/2 elements from the large cycle and convert that many 1-cycles to 2-cycles, we replace some terms in the denominator ((m-1)(m-2)(m-3)...) by terms that resemble (m-1)(m-3)(m-5)... (Or possibly (m)(m-2)(m-4)...), and n-m+1 gets replaced by something like n-3(m-1)/2.  I am not bothering to use integer arithmetic, as the messy justification that the denominator is made smaller I leave to you. In any case, I assert we can reduce the denominator by replacing half or more of the 1-cycles by 2-cycles, especially when m is at least 6.

Now we can repeat this: replace slightly more than half the 2-cycles with 3-cycles, borrowing about m/4 from the large cycle. We get another reduction, but with a factor of about (3/2)^(m/4) involved as opposed to (2/1)^(m/2). We can iterate this up to log m times about, taking half or more k cycles and making (k+1) cycles from them. With care we borrow about m elements from the large cycle, and make cycle lengths up to log m.

Do we stop there? Not if n is big enough. We can now use a 1-cycle (we have about m/2 of them) and borrow from the large cycle to create a single cycle of length slightly larger than log m. We can repeat this as long as the cycle created has length less than the number of remaining 1-cycles. Similarly, we can reuse a few of the m/4 2-cycles this way.

We do not get quite up to cycles of length sqrt(m) this way because of the repetitions of small cycles, but for large n we have made a series of reductions in the denominator. Using this as a threshold, one can try perturbing this structure slightly to improve the denominator, without having to search the whole space.

**End Update 2019.06.11.**

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