[proofplan] The proof uses only the defining trace property of $\varphi$. First handle the one-letter case separately so that no empty product convention is needed. For $k\geq 2$, group the word as $a_1(a_2\cdots a_k)$ and apply traciality to move the first factor to the end. Repeating this one-step cyclic shift gives equality for every cyclic rotation. [/proofplan] [step:Handle the one-letter case directly] If $k=1$, the asserted one-step cyclic identity is \begin{align*} \varphi(a_1)=\varphi(a_1). \end{align*} This holds by reflexivity of equality. Hence assume for the rest of the proof that $k\geq 2$. [/step] [step:Move the first letter to the end using traciality] Define $b\in\mathcal A$ by \begin{align*} b:=a_2a_3\cdots a_k. \end{align*} This element is well-defined because $\mathcal A$ is an associative algebra and $k\geq 2$. Since $\varphi$ is tracial, applying $\varphi(xy)=\varphi(yx)$ with $x=a_1$ and $y=b$ gives \begin{align*} \varphi(a_1b)=\varphi(ba_1). \end{align*} Substituting the definition of $b$ and using associativity of multiplication in $\mathcal A$, we obtain \begin{align*} \varphi(a_1a_2\cdots a_k)=\varphi(a_2\cdots a_ka_1). \end{align*} [guided] The point is to turn a product of $k$ elements into a product of two algebra elements, because traciality applies to two inputs at a time. Since $k\geq 2$, define the tail element $b\in\mathcal A$ by \begin{align*} b:=a_2a_3\cdots a_k. \end{align*} The product belongs to $\mathcal A$ because $\mathcal A$ is closed under multiplication, and associativity ensures that the notation $a_2a_3\cdots a_k$ is unambiguous once the usual left-to-right parenthesisation is suppressed. Now apply the trace property of the state $\varphi$. Traciality says that for every pair $x,y\in\mathcal A$, \begin{align*} \varphi(xy)=\varphi(yx). \end{align*} Taking $x=a_1$ and $y=b$ gives \begin{align*} \varphi(a_1b)=\varphi(ba_1). \end{align*} Finally substitute $b=a_2a_3\cdots a_k$. By associativity, $a_1b=a_1a_2\cdots a_k$ and $ba_1=a_2\cdots a_ka_1$. Therefore \begin{align*} \varphi(a_1a_2\cdots a_k)=\varphi(a_2\cdots a_ka_1). \end{align*} This proves the basic one-step cyclic shift. [/guided] [/step] [step:Iterate the one-step shift to obtain every cyclic rotation] For each $r\in\{0,1,\dots,k-1\}$, define the cyclically rotated word $W_r\in\mathcal A$ by \begin{align*} W_r:=a_{r+1}a_{r+2}\cdots a_k a_1a_2\cdots a_r, \end{align*} where the final product $a_1a_2\cdots a_r$ is omitted when $r=0$. Thus $W_0=a_1a_2\cdots a_k$. The one-step cyclic shift proved above applies to any ordered $k$-tuple of elements of $\mathcal A$. Applying it to the ordered tuple \begin{align*} (a_{r+1},a_{r+2},\dots,a_k,a_1,\dots,a_r) \end{align*} for each $r\in\{0,1,\dots,k-2\}$ gives \begin{align*} \varphi(W_r)=\varphi(W_{r+1}). \end{align*} Chaining these equalities yields, for every $r\in\{0,1,\dots,k-1\}$, \begin{align*} \varphi(W_0)=\varphi(W_r). \end{align*} Since $W_0=a_1a_2\cdots a_k$, this is exactly \begin{align*} \varphi(a_1a_2\cdots a_k)=\varphi(a_{r+1}\cdots a_k a_1\cdots a_r). \end{align*} Therefore all cyclic rotations of the word have the same moment under $\varphi$. [/step]