[proofplan] Fix the variables $a_1,\dots,a_n$ and view the partition-indexed moment and cumulant functionals as functions on the finite poset $NC(n)$. The defining free moment-cumulant relation, applied block by block, says that the moment function is the zeta transform of the cumulant function on this poset. Since $NC(n)$ is finite, the ordinary Möbius inversion theorem for finite posets applies and recovers the cumulant function from the moment function. Finally, evaluating the resulting formula at the one-block partition gives the usual formula for the scalar free cumulant $\kappa_n$. [/proofplan]

[step:Encode moments and cumulants as functions on $NC(n)$] Fix the elements $a_1,\dots,a_n \in \mathcal A$. Define the functions \begin{align*} F: NC(n) \to \mathbb C \end{align*} and \begin{align*} K: NC(n) \to \mathbb C \end{align*} by \begin{align*} F(\pi) = \varphi_\pi[a_1,\dots,a_n] \end{align*} and \begin{align*} K(\pi) = \kappa_\pi[a_1,\dots,a_n]. \end{align*} The set $NC(n)$ is finite, so every sum over an interval in $NC(n)$ is a finite sum in $\mathbb C$. [/step]

[step:Show that the moment function is the zeta transform of the cumulant function]Let $\sigma \in NC(n)$. For a block $V \in \sigma$, write its elements in increasing order as $V = \{i_1 < \dots < i_r\}$. By definition of the multiplicative moment functional, \begin{align*} \varphi_\sigma[a_1,\dots,a_n] = \prod_{V \in \sigma} \varphi(a_{i_1}\cdots a_{i_r}), \end{align*} where $V = \{i_1 < \dots < i_r\}$ in each factor of the product. For each such block $V$, the defining free moment-cumulant relation gives \begin{align*} \varphi(a_{i_1}\cdots a_{i_r}) = \sum_{\theta \in NC(r)} \kappa_\theta[a_{i_1},\dots,a_{i_r}]. \end{align*} Multiplying these identities over all blocks $V \in \sigma$ and expanding the finite product of finite sums gives \begin{align*} \varphi_\sigma[a_1,\dots,a_n] = \sum_{\pi \in NC(n):\,\pi \leq \sigma} \kappa_\pi[a_1,\dots,a_n]. \end{align*} Thus \begin{align*} F(\sigma) = \sum_{\pi \in NC(n):\,\pi \leq \sigma} K(\pi). \end{align*}[/step]

[guided]We prove the displayed identity by expanding the moment-cumulant relation separately on each block of $\sigma$. Let $\sigma \in NC(n)$, and let $V \in \sigma$ be one of its blocks. Write $V = \{i_1 < \dots < i_r\}$ for the increasing enumeration of the elements of $V$. The multiplicative moment functional is defined by taking the ordinary moment on each block and multiplying the resulting scalars, so \begin{align*} \varphi_\sigma[a_1,\dots,a_n] = \prod_{V \in \sigma} \varphi(a_{i_1}\cdots a_{i_r}), \end{align*} where in every factor $V = \{i_1 < \dots < i_r\}$. Now fix one block $V = \{i_1 < \dots < i_r\}$. The free cumulants are defined by the free moment-cumulant relation on noncrossing partitions, and in this block the relation states \begin{align*} \varphi(a_{i_1}\cdots a_{i_r}) = \sum_{\theta \in NC(r)} \kappa_\theta[a_{i_1},\dots,a_{i_r}]. \end{align*} This relation applies because the entries $a_{i_1},\dots,a_{i_r}$ are elements of the same noncommutative probability space $(\mathcal A,\varphi)$. Substituting this expansion into each block factor gives a finite product of finite sums. Expanding that product amounts to choosing, for each block $V$ of $\sigma$, a noncrossing partition of the ordered list of indices inside $V$. Such a choice is equivalent to choosing a noncrossing partition $\pi$ of $\{1,\dots,n\}$ whose every block is contained in a block of $\sigma$, which is exactly the refinement condition $\pi \leq \sigma$. Under this correspondence, the product of the chosen blockwise cumulant factors is precisely the multiplicative cumulant functional $\kappa_\pi[a_1,\dots,a_n]$. Therefore \begin{align*} \varphi_\sigma[a_1,\dots,a_n] = \sum_{\pi \in NC(n):\,\pi \leq \sigma} \kappa_\pi[a_1,\dots,a_n]. \end{align*} Using the functions $F$ and $K$ defined above, this becomes \begin{align*} F(\sigma) = \sum_{\pi \in NC(n):\,\pi \leq \sigma} K(\pi). \end{align*}[/guided]

[step:Apply Möbius inversion on the finite lattice $NC(n)$] Let $\zeta_{NC}: NC(n) \times NC(n) \to \mathbb C$ be the zeta function of the poset $NC(n)$, defined by $\zeta_{NC}(\rho,\tau)=1$ if $\rho \leq \tau$ and $\zeta_{NC}(\rho,\tau)=0$ otherwise. The preceding step says \begin{align*} F(\sigma) = \sum_{\pi \in NC(n)} \zeta_{NC}(\pi,\sigma)K(\pi). \end{align*} The Möbius function $\mu_{NC}: NC(n) \times NC(n) \to \mathbb C$ is the inverse of $\zeta_{NC}$ in the incidence algebra of the finite poset $NC(n)$. Therefore, for all $\rho,\sigma \in NC(n)$ with $\rho \leq \sigma$, \begin{align*} \sum_{\pi \in NC(n):\,\rho \leq \pi \leq \sigma} \mu_{NC}(\pi,\sigma) = 1 \end{align*} when $\rho=\sigma$, and \begin{align*} \sum_{\pi \in NC(n):\,\rho \leq \pi \leq \sigma} \mu_{NC}(\pi,\sigma) = 0 \end{align*} when $\rho<\sigma$. Fix $\sigma \in NC(n)$. Multiplying the identity for $F(\pi)$ by $\mu_{NC}(\pi,\sigma)$ and summing over $\pi \leq \sigma$ gives \begin{align*} \sum_{\pi \in NC(n):\,\pi \leq \sigma} \mu_{NC}(\pi,\sigma)F(\pi) = \sum_{\pi \in NC(n):\,\pi \leq \sigma} \mu_{NC}(\pi,\sigma) \sum_{\rho \in NC(n):\,\rho \leq \pi} K(\rho). \end{align*} Because $NC(n)$ is finite, we may interchange the two finite sums. Hence \begin{align*} \sum_{\pi \in NC(n):\,\pi \leq \sigma} \mu_{NC}(\pi,\sigma)F(\pi) = \sum_{\rho \in NC(n):\,\rho \leq \sigma} K(\rho) \sum_{\pi \in NC(n):\,\rho \leq \pi \leq \sigma} \mu_{NC}(\pi,\sigma). \end{align*} By the preceding Möbius identity, the inner sum equals $1$ for $\rho=\sigma$ and equals $0$ for $\rho<\sigma$. Thus \begin{align*} \sum_{\pi \in NC(n):\,\pi \leq \sigma} \mu_{NC}(\pi,\sigma)F(\pi) = K(\sigma). \end{align*} Substituting the definitions of $F$ and $K$ gives \begin{align*} \kappa_\sigma[a_1,\dots,a_n] = \sum_{\pi \in NC(n):\,\pi \leq \sigma} \mu_{NC}(\pi,\sigma)\,\varphi_\pi[a_1,\dots,a_n]. \end{align*} [/step]

[step:Specialize the inversion formula to the one-block partition] Let $1_n \in NC(n)$ denote the one-block partition of $\{1,\dots,n\}$. By definition of the multiplicative cumulant functional on a one-block partition, \begin{align*} \kappa_{1_n}[a_1,\dots,a_n] = \kappa_n(a_1,\dots,a_n). \end{align*} Applying the formula just proved with $\sigma = 1_n$ gives \begin{align*} \kappa_n(a_1,\dots,a_n) = \sum_{\pi \in NC(n):\,\pi \leq 1_n} \mu_{NC}(\pi,1_n)\,\varphi_\pi[a_1,\dots,a_n]. \end{align*} Every $\pi \in NC(n)$ satisfies $\pi \leq 1_n$, so the summation domain is all of $NC(n)$. Therefore \begin{align*} \kappa_n(a_1,\dots,a_n) = \sum_{\pi \in NC(n)} \mu_{NC}(\pi,1_n)\,\varphi_\pi[a_1,\dots,a_n]. \end{align*} This is the stated special case and completes the proof. [/step]