An inequality about Stirling numbers For all positive integers $n,r,k$ that $n=rk$, how to prove the inequality
$$(r-1)!^k\le \frac{c(n,k)}{S(n,k)}\le (n-k)!$$
$c(n,k)$ is unsigned Stirling numbers of the first kind and $S(n,k)$ is Stirling numbers of the second kind. https://en.wikipedia.org/wiki/Stirling_number#Notation
Any help would be appreciated!
 A: Long Hint:
So the difference in between $S(n,k)$ and $c(n,k)$ is that the later has a cyclic order attached to each block. First you can show that $$c(n,k)\leq (n-k)!\cdot S(n,k)$$ using the following injection. Let $\sigma = (a_1 \cdots)\cdots (a_k\cdots )$ be a permutation with $k$ cycles and consider $$\varphi (\sigma)=(\{\{a_1,\cdots\},\cdots ,\{a_k\cdots \}\},\sigma'),$$ where the first part of the tuple is just forgetting the cyclic order in each cycle and the second part is a permutation that saves the order of the elements. For example:
$$\varphi ((1\,8\,3)(2\,6)(4)(5\,9\, 7))=(\{1,3,8\},\{2,6\},\{4\},\{5,7,9\},132456798).$$
Notice that i could have written the big permutation as a sequence of little permutations like $(132,45,6,798).$ Notice also that the permutation $\sigma '$ is giving the order in each block to the original cycle. Show that this is injective, and so the inequality is attained.
For the other inequality take the same function but use it in permutations where each cycle has $r$ elements like
$\sigma = \underbrace{(a_1\cdots)}_{r\text{ elements}}\cdots \underbrace{(a_k\cdots)}_{r\text{ elements}}$ you get everything in (remember that the number of cyclic permutations$(r-1)!$) in $r$ elements is    $(r-1)!^kS(n,k).$ so show that this is Surjective. Notice that there are more elements in $c(n,k)$ than the ones you are using here, for example, the ones in which every cycle is not $r$ elements, so the inequality is attained.
