[proofplan]
The proof is a direct separation of the one-block term in the [moment-cumulant formula for free cumulants](/theorems/7107). The maximal partition $1_n$ contributes exactly $\kappa_n(x_1,\dots,x_n)$, while every other noncrossing partition has all blocks of size strictly smaller than $n$, so its partition cumulant involves only lower-order cumulants. Rearranging gives the displayed recursion, and induction then gives determination of cumulants from moments. The reverse determination is obtained by evaluating the moment-cumulant formula using the known cumulants.
[/proofplan]
[step:Separate the one-block term in the moment-cumulant formula]
Fix $n \in \mathbb{N}$ and $x_1,\dots,x_n \in A$. By the moment-cumulant formula for free cumulants (citing a result not yet in the wiki: Moment-cumulant formula for free cumulants), applied to the noncommutative probability space $(A,\varphi)$ and the tuple $(x_1,\dots,x_n) \in A^n$, we have
\begin{align*}
\varphi(x_1\cdots x_n)=\sum_{\pi\in NC(n)}\kappa_\pi[x_1,\dots,x_n].
\end{align*}
The partition $1_n$ is the unique partition of $\{1,\dots,n\}$ with one block, namely $\{1,\dots,n\}$. Therefore its partition cumulant is
\begin{align*}
\kappa_{1_n}[x_1,\dots,x_n]=\kappa_n(x_1,\dots,x_n).
\end{align*}
Separating this term from the sum gives
\begin{align*}
\varphi(x_1\cdots x_n)=\kappa_n(x_1,\dots,x_n)+\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n].
\end{align*}
[guided]
The starting point is the defining moment-cumulant identity for free cumulants. It says that the moment of the ordered product $x_1\cdots x_n$ is obtained by summing one partition cumulant over each noncrossing partition of $\{1,\dots,n\}$. Applied to the present tuple, the identity is
\begin{align*}
\varphi(x_1\cdots x_n)=\sum_{\pi\in NC(n)}\kappa_\pi[x_1,\dots,x_n].
\end{align*}
Here the relevant cited input is the moment-cumulant formula for free cumulants (citing a result not yet in the wiki: Moment-cumulant formula for free cumulants). Its hypotheses are exactly the present ones: $(A,\varphi)$ is a noncommutative probability space and $x_1,\dots,x_n$ are elements of $A$.
The useful observation is that the sum contains one term of top order. The maximal partition $1_n$ has a single block, namely $\{1,\dots,n\}$. By the definition of the partition cumulant associated to a one-block partition, this term is
\begin{align*}
\kappa_{1_n}[x_1,\dots,x_n]=\kappa_n(x_1,\dots,x_n).
\end{align*}
All other summands are indexed by partitions $\pi \in NC(n)$ with $\pi \ne 1_n$. Therefore the moment-cumulant formula can be rewritten by isolating the $1_n$ term:
\begin{align*}
\varphi(x_1\cdots x_n)=\kappa_n(x_1,\dots,x_n)+\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n].
\end{align*}
This is the whole mechanism behind the recursion: the only unknown top-order cumulant appears once and with coefficient $1$.
[/guided]
[/step]
[step:Show that every remaining term uses only lower-order cumulants]
Let $\pi \in NC(n)$ satisfy $\pi \ne 1_n$. Since $1_n$ is the unique one-block partition, $\pi$ has at least two blocks. If $V \in \pi$, then $V$ is a proper subset of $\{1,\dots,n\}$, so
\begin{align*}
1 \le |V| \le n-1.
\end{align*}
Writing each block as $V=\{i_1<\cdots<i_{|V|}\}$, the corresponding factor in $\kappa_\pi[x_1,\dots,x_n]$ is
\begin{align*}
\kappa_{|V|}(x_{i_1},\dots,x_{i_{|V|}}).
\end{align*}
Thus every factor appearing in $\kappa_\pi[x_1,\dots,x_n]$ is a cumulant of order strictly smaller than $n$.
[/step]
[step:Rearrange the identity to obtain the recursion]
From the separated moment-cumulant identity,
\begin{align*}
\varphi(x_1\cdots x_n)=\kappa_n(x_1,\dots,x_n)+\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n].
\end{align*}
Subtracting the finite sum over $\pi \ne 1_n$ from both sides in $\mathbb{C}$ gives
\begin{align*}
\kappa_n(x_1,\dots,x_n)=\varphi(x_1\cdots x_n)-\sum_{\substack{\pi\in NC(n):\pi\ne 1_n}}\kappa_\pi[x_1,\dots,x_n].
\end{align*}
This is the asserted recursive formula.
[/step]
[step:Deduce that moments determine cumulants by induction]
For $m \in \mathbb{N}$, define the moment functional
\begin{align*}
M_m: A^m &\to \mathbb{C}
\end{align*}
\begin{align*}
(y_1,\dots,y_m) &\mapsto \varphi(y_1\cdots y_m).
\end{align*}
We prove by induction on $m$ that $M_1,\dots,M_m$ determine $\kappa_1,\dots,\kappa_m$.
For $m=1$, the set $NC(1)$ contains only $1_1$, so the moment-cumulant formula gives
\begin{align*}
M_1(y_1)=\kappa_1(y_1)
\end{align*}
for every $y_1 \in A$. Hence $M_1$ determines $\kappa_1$.
Assume that $1 < m \le n$ and that $M_1,\dots,M_{m-1}$ determine $\kappa_1,\dots,\kappa_{m-1}$. Applying the recursion with $m$ in place of $n$, for every $y_1,\dots,y_m \in A$ we have
\begin{align*}
\kappa_m(y_1,\dots,y_m)=M_m(y_1,\dots,y_m)-\sum_{\substack{\pi\in NC(m):\pi\ne 1_m}}\kappa_\pi[y_1,\dots,y_m].
\end{align*}
By the previous step, each summand with $\pi \ne 1_m$ uses only cumulants of orders at most $m-1$. These are already determined by the induction hypothesis, and $M_m$ is known by assumption. Therefore $\kappa_m$ is determined. Induction gives that $M_1,\dots,M_n$ determine $\kappa_1,\dots,\kappa_n$.
[/step]
[step:Deduce that cumulants determine moments from the moment-cumulant formula]
Assume that the cumulant functionals $\kappa_1,\dots,\kappa_n$ are known. For each $m$ with $1 \le m \le n$ and each $y_1,\dots,y_m \in A$, the moment-cumulant formula gives
\begin{align*}
M_m(y_1,\dots,y_m)=\varphi(y_1\cdots y_m)=\sum_{\pi\in NC(m)}\kappa_\pi[y_1,\dots,y_m].
\end{align*}
Every block of every $\pi \in NC(m)$ has cardinality at most $m$, so every factor in every partition cumulant on the right-hand side is one of the known cumulants $\kappa_1,\dots,\kappa_m$. Since $m \le n$, all these cumulants are included among $\kappa_1,\dots,\kappa_n$. Thus each $M_m$ with $1 \le m \le n$ is determined, completing the proof.
[/step]
