[proofplan] We prove the identity by induction on $k$. The cases $k=0$ and $k=1$ identify the conventions: $C_0$ is the convolution identity and $C_1$ is the zeta function. For the induction step, convolution by $\zeta$ sums over the last intermediate endpoint $z$ before $y$. This sum is exactly the decomposition of a length-$(k+1)$ multichain according to its penultimate element. [/proofplan]

[step:Identify the zeroth and first powers with the corresponding multichain counts] Let $\delta\in I(P;R)$ denote the convolution identity, so $\delta(x,x)=1_R$ and $\delta(x,y)=0_R$ when $x<y$. By convention, $\zeta^0=\delta$. For $x\le y$, a multichain of length $0$ from $x$ to $y$ is a sequence \begin{align*} x=x_0=y. \end{align*} Thus there is exactly one such multichain if $x=y$, and none if $x<y$. Interpreting finite counts in $R$ through the canonical integer action on $1_R$, this gives \begin{align*} C_0(x,y)=\delta(x,y). \end{align*} Hence $\zeta^0=C_0$. For $k=1$, a multichain from $x$ to $y$ has the form \begin{align*} x=x_0\le x_1=y. \end{align*} Since $I(P;R)$ is defined only on comparable pairs $x\le y$, there is exactly one such multichain for every pair in the domain. Therefore \begin{align*} C_1(x,y)=1_R=\zeta(x,y), \end{align*} so $\zeta^1=C_1$. [/step]

[step:Compute convolution by $\zeta$ as a sum over the penultimate endpoint]Assume that $k\ge 1$ and that $\zeta^k=C_k$ in $I(P;R)$. Fix a comparable pair $x\le y$ in $P$. Since $P$ is locally finite, the interval \begin{align*} [x,y]=\{z\in P:x\le z\le y\} \end{align*} is finite, so the convolution sum is finite. Using the convolution product in the incidence algebra, as defined in [citetheorem:8100], we obtain \begin{align*} \zeta^{k+1}(x,y)=(\zeta^k*\zeta)(x,y). \end{align*} By the induction hypothesis and the definition of convolution, \begin{align*} (\zeta^k*\zeta)(x,y)=\sum_{x\le z\le y} C_k(x,z)\zeta(z,y). \end{align*} Since $\zeta(z,y)=1_R$ for every $z$ satisfying $x\le z\le y$, this becomes \begin{align*} \zeta^{k+1}(x,y)=\sum_{x\le z\le y} C_k(x,z). \end{align*}[/step]

[guided]The induction hypothesis says that $\zeta^k$ already counts length-$k$ multichains. To pass from length $k$ to length $k+1$, we multiply once more by $\zeta$ using convolution in the incidence algebra. Fix $x\le y$. Local finiteness is needed here: it guarantees that the interval \begin{align*} [x,y]=\{z\in P:x\le z\le y\} \end{align*} is finite, so the convolution sum defining $(\zeta^k*\zeta)(x,y)$ is a finite sum in $R$. By the convolution formula in $I(P;R)$, \begin{align*} (\zeta^k*\zeta)(x,y)=\sum_{x\le z\le y}\zeta^k(x,z)\zeta(z,y). \end{align*} The induction hypothesis gives $\zeta^k(x,z)=C_k(x,z)$ for every $z$ in the interval. The definition of the zeta function gives $\zeta(z,y)=1_R$ for every comparable pair $z\le y$. Substituting both facts gives \begin{align*} \zeta^{k+1}(x,y)=\sum_{x\le z\le y} C_k(x,z)1_R. \end{align*} Since multiplication by $1_R$ leaves each term unchanged, this is \begin{align*} \zeta^{k+1}(x,y)=\sum_{x\le z\le y} C_k(x,z). \end{align*} This formula is the algebraic form of the combinatorial idea: choose the penultimate endpoint $z$, count all length-$k$ multichains from $x$ to $z$, and then append the final relation $z\le y$.[/guided]

[step:Match the convolution sum with the number of length-$(k+1)$ multichains] For each $z\in [x,y]$, let $\mathcal M_k(x,z)$ be the finite set of multichains \begin{align*} x=x_0\le x_1\le \cdots \le x_k=z. \end{align*} Let $\mathcal M_{k+1}(x,y)$ be the finite set of multichains \begin{align*} x=x_0\le x_1\le \cdots \le x_k\le x_{k+1}=y. \end{align*} Define a map \begin{align*} \Phi:\bigsqcup_{z\in [x,y]}\mathcal M_k(x,z)&\to \mathcal M_{k+1}(x,y) \end{align*} \begin{align*} (z,(x_0,\dots,x_k))&\mapsto (x_0,\dots,x_k,y). \end{align*} This map is well-defined because $x_k=z\le y$. It is injective because the image records $z$ as its penultimate element $x_k$. It is surjective because every chain \begin{align*} x=x_0\le x_1\le \cdots \le x_k\le x_{k+1}=y \end{align*} has penultimate element $z=x_k$, and its initial segment \begin{align*} x=x_0\le x_1\le \cdots \le x_k=z \end{align*} lies in $\mathcal M_k(x,z)$. Hence $\Phi$ is a bijection. Therefore the finite count of $\mathcal M_{k+1}(x,y)$ is the sum of the finite counts of the sets $\mathcal M_k(x,z)$: \begin{align*} C_{k+1}(x,y)=\sum_{x\le z\le y} C_k(x,z). \end{align*} Comparing this with the convolution computation gives \begin{align*} \zeta^{k+1}(x,y)=C_{k+1}(x,y). \end{align*} Since this holds for every comparable pair $x\le y$, we have $\zeta^{k+1}=C_{k+1}$ in $I(P;R)$. By induction, $\zeta^k=C_k$ for every integer $k\ge 0$. [/step]