[proofplan] The proof is a direct use of the trace property. We split the word $x_1\cdots x_n$ into an initial block and a terminal block, then commute the two blocks inside $\varphi$. This gives equality of all cyclic rotations. Finally, applying the same identity to monomials in noncommuting indeterminates shows that the tracial joint law has redundant word-moment data along cyclic equivalence classes. [/proofplan] [step:Split the word at the chosen cyclic rotation] Fix an integer $k$ with $0 \leq k < n$. Define two elements $u_k,v_k \in A$ by \begin{align*} u_k := x_1\cdots x_k,\qquad v_k := x_{k+1}\cdots x_n, \end{align*} with the convention that $u_0=1_A$ when $k=0$. Then multiplication in $A$ gives \begin{align*} u_kv_k=x_1\cdots x_n. \end{align*} Similarly, \begin{align*} v_ku_k=x_{k+1}\cdots x_nx_1\cdots x_k. \end{align*} [/step] [step:Apply traciality to interchange the two blocks] Since $\varphi$ is tracial, it satisfies $\varphi(ab)=\varphi(ba)$ for every pair $a,b \in A$. Applying this with $a=u_k$ and $b=v_k$ gives \begin{align*} \varphi(u_kv_k)=\varphi(v_ku_k). \end{align*} Using the identities from the previous step, this becomes \begin{align*} \varphi(x_1\cdots x_n)=\varphi(x_{k+1}\cdots x_nx_1\cdots x_k). \end{align*} [guided] The only structural property used here is traciality. After fixing $k$, the word $x_1\cdots x_n$ is written as a product of two blocks: \begin{align*} u_k := x_1\cdots x_k,\qquad v_k := x_{k+1}\cdots x_n. \end{align*} When $k=0$, the first block contains no letters, so it is the empty product, which by convention is the unit $1_A$. The original word is exactly the product of the two blocks in the order $u_kv_k$: \begin{align*} u_kv_k=x_1\cdots x_n. \end{align*} The cyclically rotated word is the same pair of blocks in the reversed block order: \begin{align*} v_ku_k=x_{k+1}\cdots x_nx_1\cdots x_k. \end{align*} Because $\varphi$ is tracial, it does not distinguish $u_kv_k$ from $v_ku_k$ inside the expectation: \begin{align*} \varphi(u_kv_k)=\varphi(v_ku_k). \end{align*} Substituting the two block identities gives \begin{align*} \varphi(x_1\cdots x_n)=\varphi(x_{k+1}\cdots x_nx_1\cdots x_k). \end{align*} This proves the cyclic invariance identity for the fixed rotation parameter $k$. [/guided] [/step] [step:Recover tracial word moments from cyclic representatives] Let $r \in \mathbb{N}$ and let $x=(x_1,\dots,x_r) \in A^r$. Define the evaluation homomorphism $\operatorname{ev}_x: \mathbb{C}\langle Z_1,\dots,Z_r\rangle \to A$ by $\operatorname{ev}_x(P)=P(x_1,\dots,x_r)$. The joint law of $x$ is the linear functional $L_x=\varphi \circ \operatorname{ev}_x$. For an integer $m \in \mathbb{N}$, two words $Z_{i_1}\cdots Z_{i_m}$ and $Z_{j_1}\cdots Z_{j_m}$ are called cyclically equivalent when there is an integer $k$ with $0 \leq k < m$ such that \begin{align*} Z_{j_1}\cdots Z_{j_m}=Z_{i_{k+1}}\cdots Z_{i_m}Z_{i_1}\cdots Z_{i_k}. \end{align*} For any word $Z_{i_1}\cdots Z_{i_m}$ in the noncommuting indeterminates and any integer $k$ with $0 \leq k < m$, the identity already proved, applied to the elements $x_{i_1},\dots,x_{i_m} \in A$, gives \begin{align*} L_x(Z_{i_1}\cdots Z_{i_m})=L_x(Z_{i_{k+1}}\cdots Z_{i_m}Z_{i_1}\cdots Z_{i_k}). \end{align*} Thus word moments of $L_x$ are constant on cyclic equivalence classes. Since a noncommutative polynomial is a finite linear combination of words and $L_x$ is linear, specifying the value of $L_x$ on one word representative from each cyclic class determines all tracial word-moment values. This is precisely the asserted cyclic reduction for tracial joint laws. [/step]