[proofplan]
The element $\eta=\zeta-\delta$ is the strict-comparability function: it is $1_R$ on strict intervals and $0_R$ on diagonal intervals. We first show that $\eta$ is nilpotent, so the exponential is a finite algebraic sum rather than an analytic limit. Then we expand the convolution power $\eta^k(x,y)$ and observe that exactly the strict chains from $x$ to $y$ contribute nonzero terms. Substituting $\eta^k(x,y)=c_k(x,y)1_R$ into the finite exponential series gives the desired formula.
[/proofplan]
[step:Identify $\eta$ as the strict-comparability function]
By definition of $\zeta$ and $\delta$, for every $x\le y$ in $P$,
\begin{align*}
\eta(x,y)=\zeta(x,y)-\delta(x,y).
\end{align*}
Thus $\eta(x,x)=0_R$ for every $x\in P$, and $\eta(x,y)=1_R$ for every strict comparable pair $x<y$. Since $R$ is a $\mathbb Q$-algebra, each rational number $1/k!$ acts as a scalar in $R$, so the coefficients in the exponential series are well-defined.
[/step]
[step:Show that the exponential series is finite]
Let $n=|P|$. Since $P$ is finite, it has a linear extension by [citetheorem:8086]. Choose a bijection
\begin{align*}
\ell:P\to \{1,\dots,n\}
\end{align*}
such that $a<b$ in $P$ implies $\ell(a)<\ell(b)$. With respect to this ordering, the matrix of $\eta$ is strictly upper triangular because $\eta(a,b)\ne 0_R$ only when $a<b$. Hence every nonzero product contributing to $\eta^k(a,b)$ forces a strictly increasing sequence of $\ell$-values of length $k+1$. Such a sequence cannot exist when $k\ge n$. Therefore
\begin{align*}
\eta^k=0
\end{align*}
in $I(P;R)$ for every $k\ge n$, and consequently
\begin{align*}
\exp(\eta)=\sum_{k=0}^{n-1}\frac{\eta^k}{k!}.
\end{align*}
[guided]
Let $n=|P|$. The goal of this step is to justify that the expression called $\exp(\eta)$ is a finite algebraic sum. Since $P$ is finite, [citetheorem:8086] gives a linear extension of $P$. Thus we may choose a bijection
\begin{align*}
\ell:P\to \{1,\dots,n\}
\end{align*}
with the property that whenever $a<b$ in $P$, one has $\ell(a)<\ell(b)$.
From the previous step, $\eta(a,b)$ can be nonzero only when $a<b$. Therefore, in the ordering of $P$ determined by $\ell$, the matrix of $\eta$ has zeros on the diagonal and below the diagonal: it is strictly upper triangular. A product contributing to $\eta^k(a,b)$ has the form
\begin{align*}
\eta(x_0,x_1)\eta(x_1,x_2)\cdots \eta(x_{k-1},x_k)
\end{align*}
with $x_0=a$ and $x_k=b$. For this product to be nonzero, every factor must be nonzero, so
\begin{align*}
x_0<x_1<\cdots <x_k.
\end{align*}
Applying the linear extension $\ell$ gives
\begin{align*}
\ell(x_0)<\ell(x_1)<\cdots <\ell(x_k).
\end{align*}
This is a strictly increasing sequence of $k+1$ distinct elements of the $n$-element set $\{1,\dots,n\}$. If $k\ge n$, such a sequence cannot exist. Hence every summand in every entry of $\eta^k$ is zero when $k\ge n$, so
\begin{align*}
\eta^k=0.
\end{align*}
Thus the exponential series is actually the finite sum
\begin{align*}
\exp(\eta)=\sum_{k=0}^{n-1}\frac{\eta^k}{k!}.
\end{align*}
No analytic convergence is being used; nilpotence inside the finite incidence algebra is doing all the work.
[/guided]
[/step]
[step:Expand convolution powers as sums over intermediate sequences]
We prove that for every $k\ge 0$ and every $x\le y$ in $P$,
\begin{align*}
\eta^k(x,y)=c_k(x,y)1_R.
\end{align*}
For $k=0$, the identity element of the incidence algebra is $\delta$, so
\begin{align*}
\eta^0(x,y)=\delta(x,y).
\end{align*}
This equals $1_R$ if $x=y$ and $0_R$ if $x<y$, which is exactly $c_0(x,y)1_R$ by the convention on length-zero strict chains.
Now let $k\ge 1$. By the convolution product in the incidence algebra [citetheorem:8100], iterating convolution gives
\begin{align*}
\eta^k(x,y)=\sum_{x=x_0\le x_1\le \cdots \le x_k=y}\eta(x_0,x_1)\eta(x_1,x_2)\cdots \eta(x_{k-1},x_k).
\end{align*}
Each factor $\eta(x_{i-1},x_i)$ equals $1_R$ precisely when $x_{i-1}<x_i$, and equals $0_R$ when $x_{i-1}=x_i$. Hence a summand is $1_R$ exactly for a strict chain
\begin{align*}
x=x_0<x_1<\cdots <x_k=y,
\end{align*}
and is $0_R$ otherwise. Therefore the sum contains exactly $c_k(x,y)$ nonzero summands, each equal to $1_R$, and so
\begin{align*}
\eta^k(x,y)=c_k(x,y)1_R.
\end{align*}
[/step]
[step:Substitute the chain count into the finite exponential]
Let $x\le y$ in $P$. Since $\eta^k=0$ for all $k\ge n=|P|$, evaluating the finite exponential series at $(x,y)$ gives
\begin{align*}
\exp(\eta)(x,y)=\sum_{k=0}^{n-1}\frac{\eta^k(x,y)}{k!}.
\end{align*}
Using $\eta^k(x,y)=c_k(x,y)1_R$ for every $k\ge 0$, we obtain
\begin{align*}
\exp(\eta)(x,y)=\sum_{k=0}^{n-1}\frac{c_k(x,y)}{k!}1_R.
\end{align*}
For $k\ge n$, there is no strict chain of length $k$ in an $n$-element poset, so $c_k(x,y)=0$. Hence the last finite sum may be written as
\begin{align*}
\exp(\eta)(x,y)=\sum_{k\ge 0}\frac{c_k(x,y)}{k!}1_R.
\end{align*}
This is the stated exponential generating function for strict chains from $x$ to $y$.
[/step]
