[proofplan]
Fix $m,n \in \mathbb N$. The proof expands $\kappa_m(S_n,\dots,S_n)$ using multilinearity of the $m$-th free cumulant, producing a sum over all $m$-tuples of indices from $\{1,\dots,n\}$. Freeness eliminates every term whose index tuple contains at least two distinct values, because such terms are mixed free cumulants. The remaining diagonal terms are all equal by identical distribution, and counting those $n$ terms gives the factor $n^{1-m/2}$.
[/proofplan]
[step:Expand the cumulant by multilinearity over all index tuples]
Fix $m,n \in \mathbb N$. Let $I_n := \{1,\dots,n\}$ denote the set of summation indices. Since the $m$-th free cumulant is a multilinear map
\begin{align*}
\kappa_m: \mathcal A^m \to \mathbb C,
\end{align*}
and since
\begin{align*}
S_n = n^{-1/2}\sum_{j=1}^{n} a_j,
\end{align*}
multilinearity in each of the $m$ arguments gives
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{(i_1,\dots,i_m) \in I_n^m}\kappa_m(a_{i_1},\dots,a_{i_m}).
\end{align*}
[guided]
We first isolate the only algebraic property of free cumulants needed at this stage: multilinearity. For the fixed positive integers $m$ and $n$, the $m$-th free cumulant is a multilinear map
\begin{align*}
\kappa_m: \mathcal A^m \to \mathbb C.
\end{align*}
Define the finite index set $I_n := \{1,\dots,n\}$. The normalized sum is
\begin{align*}
S_n = n^{-1/2}\sum_{j=1}^{n} a_j.
\end{align*}
Substituting this expression into each of the $m$ slots of $\kappa_m$ and pulling out the scalar factor $n^{-1/2}$ from each slot gives the total scalar factor $n^{-m/2}$. Expanding the resulting finite multilinear expression over all independent choices of the $m$ indices gives
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{(i_1,\dots,i_m) \in I_n^m}\kappa_m(a_{i_1},\dots,a_{i_m}).
\end{align*}
This expansion is finite, so no convergence or limiting argument is involved.
[/guided]
[/step]
[step:Discard every mixed cumulant by freeness]
For each $j \in I_n$, let $\mathcal A_j \subset \mathcal A$ denote the unital subalgebra generated by $a_j$. The assumption that $(a_j)_{j \in \mathbb N}$ are freely independent means that the subalgebras $(\mathcal A_j)_{j \in \mathbb N}$ are free. If an index tuple $(i_1,\dots,i_m) \in I_n^m$ has at least two distinct entries, then the arguments $a_{i_1},\dots,a_{i_m}$ come from at least two distinct free subalgebras. Therefore the term $\kappa_m(a_{i_1},\dots,a_{i_m})$ is a mixed free cumulant, so it vanishes by the standard vanishing theorem for mixed free cumulants of free subalgebras (citing a result not yet in the wiki: Mixed free cumulants vanish for free subalgebras).
Thus only the diagonal index tuples $(j,\dots,j)$ with $j \in I_n$ remain, and the expansion reduces to
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{j=1}^{n}\kappa_m(a_j,\dots,a_j).
\end{align*}
[/step]
[step:Identify the surviving diagonal cumulants by identical distribution]
Because the variables $(a_j)_{j \in \mathbb N}$ are identically distributed, the joint law of a single repeated variable $a_j$ agrees with that of $a_1$ for every $j \in I_n$. Free cumulants are determined by the moments through the moment-cumulant relations, so for each $j \in I_n$,
\begin{align*}
\kappa_m(a_j,\dots,a_j) = \kappa_m(a_1,\dots,a_1).
\end{align*}
Substituting this equality into the reduced diagonal sum gives
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2}\sum_{j=1}^{n}\kappa_m(a_1,\dots,a_1).
\end{align*}
Since the summand is independent of $j$, the finite sum contains exactly $n$ equal terms:
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{-m/2} n \kappa_m(a_1,\dots,a_1).
\end{align*}
[/step]
[step:Collect the power of $n$ to obtain the scaling law]
Combining the scalar factors gives
\begin{align*}
n^{-m/2} n = n^{1-m/2}.
\end{align*}
Therefore
\begin{align*}
\kappa_m(S_n,\dots,S_n) = n^{1-m/2}\kappa_m(a_1,\dots,a_1).
\end{align*}
This is the claimed scaling formula for every $m,n \in \mathbb N$.
[/step]
