[proofplan] The first cumulant is the expectation, so centering subtracts exactly the scalar $\lambda$ from $\kappa_1$. For higher cumulants, we first prove from the moment-cumulant formula that every free cumulant of order at least two vanishes whenever one of its entries is the unit. Multilinearity then expands $\kappa_n(a-\lambda 1,\dots,a-\lambda 1)$ into a sum indexed by choices of $a$ or $1$ in each slot, and all terms containing at least one $1$ vanish for $n\ge 2$. The unique surviving term is the one in which every argument is $a$. [/proofplan]

[step:Compute the first cumulant from the unital expectation] By the defining moment-cumulant formula in order $1$, the first free cumulant satisfies \begin{align*} \kappa_1(x)=\varphi(x) \end{align*} for every $x\in\mathcal A$. Hence, using linearity of $\varphi$ and the normalization $\varphi(1)=1$, \begin{align*} \kappa_1(a-\lambda 1)=\varphi(a-\lambda 1)=\varphi(a)-\lambda\varphi(1)=\varphi(a)-\lambda=\kappa_1(a)-\lambda. \end{align*} [/step]

[step:Show that higher free cumulants vanish when one argument is the unit]We prove the following claim. [claim:Unit entries kill higher free cumulants] Let $m\ge 2$, let $x_1,\dots,x_m\in\mathcal A$, and suppose $x_j=1$ for some $j\in\{1,\dots,m\}$. Then \begin{align*} \kappa_m(x_1,\dots,x_m)=0. \end{align*} [/claim] [proof] For $r\in\mathbb N$, let $NC(r)$ denote the set of noncrossing partitions of $\{1,\dots,r\}$. If $\pi\in NC(r)$, define the block-product map $\kappa_\pi:\mathcal A^r\to\mathbb C$ on $(y_1,\dots,y_r)\in\mathcal A^r$ by \begin{align*} \kappa_\pi(y_1,\dots,y_r)=\prod_{V\in\pi}\kappa_{|V|}(y_{i_1},\dots,y_{i_{|V|}}) \end{align*} where each block $V=\{i_1<\cdots<i_{|V|}\}$ is written in increasing order. We prove the claim by induction on $m$. Fix $m\ge 2$, assume the claim is known for all orders $2,\dots,m-1$, and take $x_1,\dots,x_m\in\mathcal A$ with $x_j=1$. Let $z_1,\dots,z_{m-1}\in\mathcal A$ be the list obtained from $x_1,\dots,x_m$ by deleting the $j$-th entry. Since $x_j=1$ and $1$ is the multiplicative identity of $\mathcal A$, \begin{align*} \varphi(x_1\cdots x_m)=\varphi(z_1\cdots z_{m-1}). \end{align*} Apply the moment-cumulant formula to both sides. The formula for the right-hand side gives \begin{align*} \varphi(z_1\cdots z_{m-1})=\sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}]. \end{align*} For the left-hand side, \begin{align*} \varphi(x_1\cdots x_m)=\sum_{\pi\in NC(m)}\kappa_\pi[x_1,\dots,x_m]. \end{align*} The partitions $\pi\in NC(m)$ for which $\{j\}$ is a singleton block are in bijection with $NC(m-1)$ by deleting the singleton block and relabelling the remaining ordered positions. For such a partition, the singleton contributes \begin{align*} \kappa_1(1)=\varphi(1)=1, \end{align*} so the total contribution of all partitions with singleton block $\{j\}$ is exactly \begin{align*} \sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}]. \end{align*} Now consider a partition $\pi\in NC(m)$ in which $j$ belongs to a block $V$ with $|V|\ge 2$. If $V\ne\{1,\dots,m\}$, then the factor attached to $V$ is a cumulant of order $|V|$ with one argument equal to $1$ and with $2\le |V|\le m-1$; it is zero by the induction hypothesis. Hence every such partition contributes zero. The only remaining partition of this kind is the one-block partition $\{1,\dots,m\}$, whose contribution is \begin{align*} \kappa_m(x_1,\dots,x_m). \end{align*} Combining the preceding decomposition of the moment-cumulant sum gives \begin{align*} \varphi(x_1\cdots x_m)=\sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}]+\kappa_m(x_1,\dots,x_m). \end{align*} Comparing this identity with the already established equality \begin{align*} \varphi(x_1\cdots x_m)=\varphi(z_1\cdots z_{m-1})=\sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}] \end{align*} yields \begin{align*} \kappa_m(x_1,\dots,x_m)=0. \end{align*} This completes the induction and proves the claim. [/proof][/step]

[guided]We prove the vanishing statement directly from the definition of free cumulants, rather than treating it as a black box. For $r\in\mathbb N$, let $NC(r)$ be the set of noncrossing partitions of $\{1,\dots,r\}$. If $\pi\in NC(r)$, define the block-product map $\kappa_\pi:\mathcal A^r\to\mathbb C$ on $(y_1,\dots,y_r)\in\mathcal A^r$ by \begin{align*} \kappa_\pi(y_1,\dots,y_r)=\prod_{V\in\pi}\kappa_{|V|}(y_{i_1},\dots,y_{i_{|V|}}) \end{align*} where each block $V=\{i_1<\cdots<i_{|V|}\}$ is read in increasing order. We prove by induction on $m\ge 2$ that if one argument of $\kappa_m$ is the unit, then the cumulant is zero. Fix $m\ge 2$ and suppose $x_1,\dots,x_m\in\mathcal A$ with $x_j=1$ for some $j$. Let $z_1,\dots,z_{m-1}$ be the list obtained by deleting this unit entry. The key observation is that the ordinary moment does not change when the unit is deleted: \begin{align*} \varphi(x_1\cdots x_m)=\varphi(z_1\cdots z_{m-1}). \end{align*} Now expand both sides by the moment-cumulant formula. The shorter moment is \begin{align*} \varphi(z_1\cdots z_{m-1})=\sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}]. \end{align*} The longer moment is \begin{align*} \varphi(x_1\cdots x_m)=\sum_{\pi\in NC(m)}\kappa_\pi[x_1,\dots,x_m]. \end{align*} We separate the partitions in $NC(m)$ according to what happens to the position $j$. If $\{j\}$ is a singleton block, then deleting that singleton block gives a noncrossing partition of the remaining $m-1$ ordered positions. The singleton contributes \begin{align*} \kappa_1(1)=\varphi(1)=1. \end{align*} Therefore the total contribution of these singleton partitions is exactly the moment-cumulant expansion of the shorter moment: \begin{align*} \sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}]. \end{align*} It remains to understand partitions where the unit belongs to a block of size at least two. If that block is not the whole set $\{1,\dots,m\}$, then the corresponding factor is a cumulant of some order between $2$ and $m-1$ with one unit entry. By the induction hypothesis, that factor is zero, so the entire product attached to the partition is zero. The only partition not covered by this induction argument is the one-block partition, and its contribution is precisely \begin{align*} \kappa_m(x_1,\dots,x_m). \end{align*} Thus the expansion of the longer moment reduces to \begin{align*} \varphi(x_1\cdots x_m)=\sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}]+\kappa_m(x_1,\dots,x_m). \end{align*} But the same moment is equal to the shorter moment, whose expansion is \begin{align*} \varphi(z_1\cdots z_{m-1})=\sum_{\sigma\in NC(m-1)}\kappa_\sigma[z_1,\dots,z_{m-1}]. \end{align*} Subtracting these equal complex numbers gives \begin{align*} \kappa_m(x_1,\dots,x_m)=0. \end{align*} This proves the induction step, and hence every higher free cumulant with a unit entry vanishes.[/guided]

[step:Expand the centered cumulant by multilinearity and discard the unit terms] Fix $n\ge 2$. For each subset $S\subset\{1,\dots,n\}$ and each $i\in\{1,\dots,n\}$, define $y_{S,i}\in\mathcal A$ by setting $y_{S,i}=a$ if $i\in S$ and $y_{S,i}=1$ if $i\notin S$. The free cumulant functional $\kappa_n:\mathcal A^n\to\mathbb C$ is multilinear because it is obtained from the multilinear moment functionals by Möbius inversion on the finite lattice $NC(n)$. Hence \begin{align*} \kappa_n(a-\lambda 1,\dots,a-\lambda 1)=\sum_{S\subset\{1,\dots,n\}}(-\lambda)^{n-|S|}\kappa_n(y_{S,1},\dots,y_{S,n}). \end{align*} If $S\ne\{1,\dots,n\}$, then at least one argument $y_{S,i}$ is equal to $1$, so the previous step gives \begin{align*} \kappa_n(y_{S,1},\dots,y_{S,n})=0. \end{align*} The only nonzero term is the term $S=\{1,\dots,n\}$, where $y_{S,i}=a$ for every $i$. Therefore \begin{align*} \kappa_n(a-\lambda 1,\dots,a-\lambda 1)=\kappa_n(a,\dots,a). \end{align*} Together with the first-cumulant computation, this proves the theorem. [/step]