[proofplan]
We count maximal chains in a fixed interval $[x,y]$ of rank $n$ in two ways. First, the definition of a binomial poset says that there are $B(n)$ such chains. Second, every maximal chain has a unique element at relative rank $k$, and fixing that element $z$ splits the chain into a maximal chain in $[x,z]$ and a maximal chain in $[z,y]$. Comparing the two counts gives the formula for the number of possible elements $z$ at relative rank $k$.
[/proofplan]
[step:Fix the interval and the relative rank to be counted]
Let $[x,y]\subset P$ be an interval with $\rho(y)-\rho(x)=n$, and fix an integer $k$ with $0\le k\le n$. Define
\begin{align*}
A_k=\{z\in [x,y]:\rho(z)-\rho(x)=k\}.
\end{align*}
We must prove that
\begin{align*}
|A_k|=\frac{B(n)}{B(k)B(n-k)}.
\end{align*}
Since $P$ is a binomial poset, every interval of rank $m$ has exactly $B(m)$ maximal chains, for each $m\in\mathbb{N}_0$.
[/step]
[step:Partition maximal chains by their unique element of relative rank $k$]
Let $\mathcal{M}(x,y)$ denote the finite set of maximal chains from $x$ to $y$. Thus an element of $\mathcal{M}(x,y)$ is a saturated chain
\begin{align*}
x=c_0\lessdot c_1\lessdot \cdots \lessdot c_n=y.
\end{align*}
Because the chain is saturated and has length $n$, each $c_i$ has relative rank $i$, so $\rho(c_i)-\rho(x)=i$. Therefore every chain in $\mathcal{M}(x,y)$ contains exactly one element of $A_k$, namely $c_k$.
For each $z\in A_k$, define
\begin{align*}
\mathcal{M}_z=\{C\in \mathcal{M}(x,y):z\in C\}.
\end{align*}
The sets $\mathcal{M}_z$, indexed by $z\in A_k$, are pairwise disjoint and their union is all of $\mathcal{M}(x,y)$. Hence
\begin{align*}
|\mathcal{M}(x,y)|=\sum_{z\in A_k}|\mathcal{M}_z|.
\end{align*}
[guided]
We need to count all maximal chains from $x$ to $y$, but we want the answer to reveal how many elements occur at relative rank $k$. The useful bookkeeping device is to sort each maximal chain according to the element it passes through at that relative rank.
Let $\mathcal{M}(x,y)$ be the set of maximal chains from $x$ to $y$. Since $[x,y]$ has rank $n$, every maximal chain from $x$ to $y$ has the form
\begin{align*}
x=c_0\lessdot c_1\lessdot \cdots \lessdot c_n=y.
\end{align*}
The chain is saturated, so moving from $c_i$ to $c_{i+1}$ increases the rank by $1$. Since $\rho(c_0)=\rho(x)$, it follows that
\begin{align*}
\rho(c_i)-\rho(x)=i.
\end{align*}
In particular, the element $c_k$ is the unique element of this chain whose relative rank over $x$ is $k$.
For each $z\in A_k$, define $\mathcal{M}_z$ to be the set of maximal chains from $x$ to $y$ that contain $z$. The uniqueness just proved implies that a maximal chain belongs to exactly one of these sets: it belongs to the set indexed by its unique relative-rank-$k$ element. Thus the family $(\mathcal{M}_z)_{z\in A_k}$ is a partition of $\mathcal{M}(x,y)$, and therefore
\begin{align*}
|\mathcal{M}(x,y)|=\sum_{z\in A_k}|\mathcal{M}_z|.
\end{align*}
[/guided]
[/step]
[step:Count the chains passing through a fixed relative-rank element]
Fix $z\in A_k$. The interval $[x,z]$ has rank $k$, because $\rho(z)-\rho(x)=k$. The interval $[z,y]$ has rank $n-k$, because
\begin{align*}
\rho(y)-\rho(z)=\rho(y)-\rho(x)-(\rho(z)-\rho(x))=n-k.
\end{align*}
A maximal chain from $x$ to $y$ passing through $z$ is uniquely obtained by choosing a maximal chain from $x$ to $z$ and a maximal chain from $z$ to $y$, then concatenating them at $z$. Conversely, such a concatenation is a maximal chain from $x$ to $y$ passing through $z$.
Since $P$ is binomial, there are $B(k)$ maximal chains in $[x,z]$ and $B(n-k)$ maximal chains in $[z,y]$. Thus
\begin{align*}
|\mathcal{M}_z|=B(k)B(n-k).
\end{align*}
[/step]
[step:Compare the two chain counts and identify the binomial coefficient]
By the definition of the factorial function of a binomial poset, the interval $[x,y]$ has exactly $B(n)$ maximal chains, so
\begin{align*}
|\mathcal{M}(x,y)|=B(n).
\end{align*}
Using the partition from the previous step and the fixed-$z$ count, we get
\begin{align*}
B(n)=\sum_{z\in A_k}B(k)B(n-k).
\end{align*}
Since the summand is independent of $z$, this becomes
\begin{align*}
B(n)=|A_k|B(k)B(n-k).
\end{align*}
Because $B(k)$ and $B(n-k)$ are positive integers, we may divide by their product and obtain
\begin{align*}
|A_k|=\frac{B(n)}{B(k)B(n-k)}=\binom{n}{k}_P.
\end{align*}
Rearranging gives
\begin{align*}
B(n)=\binom{n}{k}_P B(k)B(n-k).
\end{align*}
This proves both asserted identities, including the endpoint cases $k=0$ and $k=n$ under the convention $B(0)=1$.
[/step]
