Maximise $(s!)^\frac{n}{s}\frac{n}{s}!$ Let $n>0$ be a composite integer.
I want to maximise $f(s,n)=(s!)^\frac{n}{s}(\frac{n}{s})!$ over all integers $s\mid n$ with $1<s<n$. 
The rest of this post is not necessary to read, but gives background to those who like it.
Work so far:
Experimentation suggests that taking $s$ as large as possible (e.g. $n/2$ if $n$ is even) suffices, but proving this seems tricky, especially as $f(s,n)$ is not necessarily increasing with $s$ (e.g. $f(2,1000)\approx 10^{1200}$, $f(5,1000)\approx 10^{800}$).
I tried to maximise $\log(f(s,n))=\frac{n}{s}\sum_{i=1}^s\log i\,\,\,+\,\,\sum_{i=1}^\frac{n}{s}\log i$ using integrals to bound the sums, but this got very messy (so I won't include it here) and I can't quite get this to work.
If anyone can make this work or has a better strategy, it would be much appreciated.
Motivation:
Let $G$ be imprimitive subgroup of $S_n$ with a block of order $s$. It is well-known that $G$ embeds into $S_s\wr S_{n/s}$. $f(s,n)$ is the order of the wreath product $S_s\wr S_{n/s}$. Therefore a maximal value of $f(s,n)$ is a maximal order of an imprimitive group of degree $n$. 
 A: Write $r = \frac{n}{s}$ so we want to maximize $g(r, n) = \left( \frac{n}{r} \right)!^r r!$. Stirling's approximation gives
$$\log g(r, n) \approx \left( n \log n - n \log r - n \right) + (r \log r - r)$$
and differentiating with respect to $r$ (so here we're solving a relaxation of the problem where we allow $r$ to be real) gives
$$- \frac{n}{r} + \log r$$
from which we conclude that $g$ starts out a decreasing function of $r$ and approaches a local minimum around where $n = r \log r$, which gives $r \approx \frac{n}{\log n}$; it increases after this. This is the only critical point, and we want to take a maximum, so $r$ should be either as large or as small as possible.
Let $p$ be the smallest prime divisor of $n$. Then we only need to consider the cases $r = p$ and $r = \frac{n}{p}$, which corresponds to $s = p$; these are the largest and smallest possibilities. This means we should also write down Stirling's approximation for $\log f(s, n)$, which we'll want to evaluate at $s = p$, which is
$$\log f(s, n) \approx (n \log s - n) + \left( \frac{n}{s} \log \frac{n}{s} - \frac{n}{s} \right).$$
If $p$ is fixed and $n \to \infty$ then the leading term of $\log g(p, n)$ is $n \log n$ while the leading term of $\log f(p, n)$ is $\frac{n}{p} \log n$ so $g$ dominates and we want to pick $r = p$. If $p \approx C \log n$ then the leading term of $\log g(p, n)$ is still $n \log n$ while the leading term of $\log f(p, n)$ is $n \log \log n$ so $g$ still dominates. And if $p$ is larger than this then both choices $r = p$ and $s = p$ occur before the local minimum $r \approx \frac{n}{\log n}$, so stay in the region where $g$ is a decreasing function of $r$; this means we should always pick $r = p$. 
