[proofplan] We compare finite-partition entropy rates for $T$ and for the iterate $T^k$. The upper bound follows because the $n$-block join for $T^k$ samples only every $k$-th time and is therefore coarser than the full $kn$-block join for $T$. For the lower bound, enlarge an arbitrary partition $\mathcal P$ to the $k$-step partition $\mathcal Q=\bigvee_{i=0}^{k-1}T^{-i}\mathcal P$; then the $n$-block join of $\mathcal Q$ under $T^k$ is exactly the $kn$-block join of $\mathcal P$ under $T$. Taking entropy rates and then suprema over finite partitions gives equality. [/proofplan]

[step:Fix the entropy notation for finite partitions] Let $\mathcal P$ be a finite measurable partition of $X$. For $m\in\mathbb N$, define the $m$-block join of $\mathcal P$ under $T$ by \begin{align*} \mathcal P_m(T):=\bigvee_{r=0}^{m-1}T^{-r}\mathcal P. \end{align*} For a finite measurable partition $\mathcal A$, write \begin{align*} H_\mu(\mathcal A):=-\sum_{A\in\mathcal A}\mu(A)\log\mu(A), \end{align*} with the convention $0\log 0=0$. The partition entropy rate is \begin{align*} h_\mu(T,\mathcal P):=\lim_{m\to\infty}\frac{1}{m}H_\mu(\mathcal P_m(T)), \end{align*} and the Kolmogorov-Sinai entropy is \begin{align*} h_\mu(T):=\sup_{\mathcal P}h_\mu(T,\mathcal P), \end{align*} where the supremum is over all finite measurable partitions $\mathcal P$ of $X$. [/step]

[step:Bound the entropy of every partition under $T^k$ from above]Fix a finite measurable partition $\mathcal P$ of $X$. For $n\in\mathbb N$, the $n$-block join of $\mathcal P$ under $T^k$ is \begin{align*} \mathcal P_n(T^k)=\bigvee_{j=0}^{n-1}(T^k)^{-j}\mathcal P=\bigvee_{j=0}^{n-1}T^{-kj}\mathcal P. \end{align*} This join uses only the times $0,k,2k,\dots,k(n-1)$, while $\mathcal P_{kn}(T)$ uses every time $0,1,\dots,kn-1$. Hence $\mathcal P_{kn}(T)$ refines $\mathcal P_n(T^k)$. Since entropy is monotone under refinement of finite partitions, \begin{align*} H_\mu(\mathcal P_n(T^k))\le H_\mu(\mathcal P_{kn}(T)). \end{align*} Dividing by $n$ gives \begin{align*} \frac{1}{n}H_\mu(\mathcal P_n(T^k))\le k\,\frac{1}{kn}H_\mu(\mathcal P_{kn}(T)). \end{align*} Taking $n\to\infty$ yields \begin{align*} h_\mu(T^k,\mathcal P)\le k\,h_\mu(T,\mathcal P)\le k\,h_\mu(T). \end{align*} Taking the supremum over all finite measurable partitions $\mathcal P$ gives \begin{align*} h_\mu(T^k)\le k\,h_\mu(T). \end{align*}[/step]

[guided]Fix a finite measurable partition $\mathcal P$ of $X$. We first compare the information recorded by observing $\mathcal P$ along the orbit of $T^k$ with the information recorded by observing $\mathcal P$ along the orbit of $T$. For $n\in\mathbb N$, the $n$-block join under $T^k$ is \begin{align*} \mathcal P_n(T^k)=\bigvee_{j=0}^{n-1}(T^k)^{-j}\mathcal P=\bigvee_{j=0}^{n-1}T^{-kj}\mathcal P. \end{align*} Thus this partition records the name of a point only at the times $0,k,2k,\dots,k(n-1)$. By contrast, the $kn$-block join under $T$ is \begin{align*} \mathcal P_{kn}(T)=\bigvee_{r=0}^{kn-1}T^{-r}\mathcal P, \end{align*} which records the name at every time $0,1,\dots,kn-1$. Knowing the full $kn$-block name determines the sparse $n$-block name, so $\mathcal P_{kn}(T)$ refines $\mathcal P_n(T^k)$. Entropy of finite partitions is monotone under refinement: if $\mathcal A$ refines $\mathcal B$, then $H_\mu(\mathcal B)\le H_\mu(\mathcal A)$. Applying this with $\mathcal A=\mathcal P_{kn}(T)$ and $\mathcal B=\mathcal P_n(T^k)$ gives \begin{align*} H_\mu(\mathcal P_n(T^k))\le H_\mu(\mathcal P_{kn}(T)). \end{align*} After dividing by $n$, we rewrite the right-hand side in terms of the $T$-entropy rate: \begin{align*} \frac{1}{n}H_\mu(\mathcal P_n(T^k))\le k\,\frac{1}{kn}H_\mu(\mathcal P_{kn}(T)). \end{align*} Letting $n\to\infty$, the left-hand side converges to $h_\mu(T^k,\mathcal P)$, while the right-hand side converges to $k\,h_\mu(T,\mathcal P)$. Therefore \begin{align*} h_\mu(T^k,\mathcal P)\le k\,h_\mu(T,\mathcal P)\le k\,h_\mu(T). \end{align*} Since this holds for every finite measurable partition $\mathcal P$, taking the supremum over $\mathcal P$ gives \begin{align*} h_\mu(T^k)\le k\,h_\mu(T). \end{align*}[/guided]

[step:Encode every $k$ consecutive $T$-names into one $T^k$-name] Fix a finite measurable partition $\mathcal P$ of $X$, and define the finite measurable partition \begin{align*} \mathcal Q:=\bigvee_{i=0}^{k-1}T^{-i}\mathcal P. \end{align*} For each $n\in\mathbb N$, \begin{align*} \mathcal Q_n(T^k)=\bigvee_{j=0}^{n-1}(T^k)^{-j}\mathcal Q. \end{align*} Using the definition of $\mathcal Q$ and $(T^k)^{-j}=T^{-kj}$, we obtain \begin{align*} \mathcal Q_n(T^k)=\bigvee_{j=0}^{n-1}\bigvee_{i=0}^{k-1}T^{-(kj+i)}\mathcal P. \end{align*} As $j$ ranges over $\{0,\dots,n-1\}$ and $i$ ranges over $\{0,\dots,k-1\}$, the integers $kj+i$ are exactly $0,1,\dots,kn-1$. Hence \begin{align*} \mathcal Q_n(T^k)=\bigvee_{r=0}^{kn-1}T^{-r}\mathcal P=\mathcal P_{kn}(T). \end{align*} Therefore \begin{align*} h_\mu(T^k,\mathcal Q)=\lim_{n\to\infty}\frac{1}{n}H_\mu(\mathcal Q_n(T^k))=k\,h_\mu(T,\mathcal P). \end{align*} Since $\mathcal Q$ is one of the finite partitions over which $h_\mu(T^k)$ takes its supremum, \begin{align*} k\,h_\mu(T,\mathcal P)=h_\mu(T^k,\mathcal Q)\le h_\mu(T^k). \end{align*} Taking the supremum over all finite measurable partitions $\mathcal P$ gives \begin{align*} k\,h_\mu(T)\le h_\mu(T^k). \end{align*} [/step]

[step:Combine the two inequalities] The preceding two steps show \begin{align*} h_\mu(T^k)\le k\,h_\mu(T) \end{align*} and \begin{align*} k\,h_\mu(T)\le h_\mu(T^k). \end{align*} Thus \begin{align*} h_\mu(T^k)=k\,h_\mu(T). \end{align*} This proves the Abramov formula for the $k$-th power of $T$. [/step]