I'm self-studying Bollobás' Combinatorics textbook and I am stuck on a particular question on Sperner families. We fix $k\ge 1$ and we know that the Sperner family ${\cal F}$ on the set $X = [n]$ contains at least one set of size at most $k$ and one set of size at least $n-k$. There are no sets in ${\cal F}$ of size strictly between $k$ and $n-k$. We want to place a bound on the size of ${\cal F}$ and in particular, the book asks us to prove that $|{\cal F}| \leq c_k n^{k-1}$ for some constant $c_k$ depending on $k$.
In the poset with all the forbidden level sets removed, no level set has more than ${n\choose k}$ elements, so we have an easy bound of $|{\cal F}| \leq {n\choose k}$. But this is $\geq n^k/k^k$, so I think this bound is too loose. My second thought was to fix a choice for the set of size $\geq n-k$, say $B = \{1,2,\ldots, n-2\}$, and then consider the poset on the first $k$ level sets that consists of sets not contained in $B$. If this poset is regularly connected, then no antichain can have more than the size of the largest level set. But I'm fumbling with the computation and can't seem to show that the largest level set in this poset has size $\sim n^{k-1}$, or even that it's regularly connected (it might well not be the case).
Most of the exercises up until now have had nice clean arguments, so I would appreciate if someone could point me in the right direction.
We can bound the elements of size at most $k$ by using a subset $A$ of size at least $n-k$ of the family.
Each of these elements can be obtained in the following way:
First select a non-empty subset of the elements that are not in $A$ (there is at most $2^k$ ways to do this).
Now select at most $k-1$ elements from the elements that are in $A$(there is at most $n^{k-1}+n^{k-1} + \dots n+1 \leq (k+1)n^{k+1}$ ways to do this).
This shows there are at most $(k+1)n^{k-1}2^k$ elements of size at most $k$ and the proof there are at most $(k+1)n^{k-1}2^k$ elements of size at least $n-k$ is analogous.