[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*}[/step]

[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: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]