[proofplan] We first identify a standard tableau of rectangular shape with a maximal chain of order ideals in the corresponding finite rectangle poset. Haiman's chain formulation of partial evacuation then identifies Schützenberger promotion with a fixed composite of two partial-evacuation operators under this tableau-chain bijection. The essential rectangular input is Haiman's rectangular dual-equivalence periodicity theorem, which says that this composite has period dividing the number of boxes on maximal chains for a rectangle. Transporting that chain identity back through the bijection gives $\operatorname{pr}^N(T)=T$. [/proofplan]

[step:Encode the tableau as a maximal chain of order ideals]Let \begin{align*} P:=\{1,\dots,a\}\times\{1,\dots,b\} \end{align*} be the rectangle poset with partial order defined by \begin{align*} (i,j)\le_P(i',j') \iff i\le i' \text{ and } j\le j'. \end{align*} An order ideal in $P$ is a subset $I\subset P$ such that $x\in I$ and $y\le_P x$ imply $y\in I$. Let $\mathcal C(P)$ denote the set of maximal chains of order ideals \begin{align*} \varnothing=I_0\subset I_1\subset\cdots\subset I_N=P \end{align*} with $|I_k|=k$ for every $0\le k\le N$. Define a map \begin{align*} \Phi:\operatorname{SYT}(\lambda)&\to\mathcal C(P) \end{align*} as follows: if $T\in\operatorname{SYT}(\lambda)$, set \begin{align*} \Phi(T)_k:=\{(i,j)\in P:T(i,j)\le k\} \end{align*} for every $0\le k\le N$. Since entries in a standard Young tableau increase strictly along rows and columns, each $\Phi(T)_k$ is an order ideal. Since exactly one box has each entry $1,\dots,N$, the sequence $\Phi(T)$ is a maximal chain. Conversely, if \begin{align*} C=(\varnothing=I_0\subset I_1\subset\cdots\subset I_N=P)\in\mathcal C(P), \end{align*} then the unique box in $I_k\setminus I_{k-1}$ receives the entry $k$. The order-ideal property forces entries to increase strictly along rows and columns, so this construction gives an element of $\operatorname{SYT}(\lambda)$. These two constructions are inverse to each other, hence $\Phi$ is a bijection.[/step]

[guided]The first point is that a standard tableau records the order in which the boxes of the rectangle are added. We use the rectangle poset \begin{align*} P:=\{1,\dots,a\}\times\{1,\dots,b\} \end{align*} with \begin{align*} (i,j)\le_P(i',j') \iff i\le i' \text{ and } j\le j'. \end{align*} This order says that a box can appear only after all boxes weakly northwest of it have appeared. For a tableau $T\in\operatorname{SYT}(\lambda)$, define \begin{align*} \Phi(T)_k:=\{(i,j)\in P:T(i,j)\le k\} \end{align*} for each $0\le k\le N$. We verify that this is an order ideal. Suppose $(i,j)\in\Phi(T)_k$ and $(i',j')\le_P(i,j)$. Then $i'\le i$ and $j'\le j$. Because $T$ is standard, entries strictly increase when moving right and down, so \begin{align*} T(i',j')\le T(i,j)\le k. \end{align*} Thus $(i',j')\in\Phi(T)_k$. Also, exactly one entry is equal to $k$ for each $1\le k\le N$, so $|\Phi(T)_k|=k$ and the sets form a maximal chain \begin{align*} \varnothing=\Phi(T)_0\subset\Phi(T)_1\subset\cdots\subset\Phi(T)_N=P. \end{align*} Conversely, start with a maximal chain \begin{align*} C=(\varnothing=I_0\subset I_1\subset\cdots\subset I_N=P). \end{align*} Since $|I_k|=k$, the difference $I_k\setminus I_{k-1}$ consists of exactly one box. Put the entry $k$ in that box. If a box $(i,j)$ is weakly southeast of another box $(i',j')$, then $(i',j')\le_P(i,j)$, and the order-ideal property implies that $(i',j')$ must enter the chain no later than $(i,j)$. Hence the assigned entries increase strictly along rows and columns. This gives a standard Young tableau, and the construction reverses the definition of $\Phi$. Therefore $\Phi$ is a bijection.[/guided]

[step:Invoke Haiman's chain identity for promotion]For each integer $m$ with $1\le m\le N$, let \begin{align*} \varepsilon_m:\mathcal C(P)\to\mathcal C(P) \end{align*} denote the $m$th partial evacuation operator on maximal chains, as defined in Haiman's dual-equivalence theory of chains of shapes. We do not use an informal reconstruction of this operator; the precise external input is Haiman's promotion-partial-evacuation identity: for every finite poset $Q$ with $N$ elements, after identifying standard tableaux or linear-extension tableaux with maximal chains of order ideals in $Q$, Schützenberger promotion is conjugate to the composite of partial evacuations \begin{align*} \varepsilon_N\circ\varepsilon_{N-1}:\mathcal C(Q)\to\mathcal C(Q). \end{align*} Applying this theorem to the finite rectangle poset $Q=P$ is legitimate because $P$ has exactly $N=ab$ elements and $\Phi$ is the tableau-chain bijection constructed above. With composition read from right to left, the theorem gives \begin{align*} \Phi\circ\operatorname{pr}\circ\Phi^{-1}=\varepsilon_N\circ\varepsilon_{N-1}. \end{align*} Equivalently, for every $T\in\operatorname{SYT}(\lambda)$, \begin{align*} \Phi(\operatorname{pr}(T))=(\varepsilon_N\circ\varepsilon_{N-1})(\Phi(T)). \end{align*} External result used: Haiman's promotion-partial-evacuation identity for maximal chains in a finite poset.[/step]

[guided]The proof now needs to translate the tableau operation $\operatorname{pr}$ into an operation on maximal chains. For each integer $m$ with $1\le m\le N$, Haiman's dual-equivalence theory defines a partial evacuation operator \begin{align*} \varepsilon_m:\mathcal C(P)\to\mathcal C(P) \end{align*} on maximal chains of order ideals. The definition is not being reconstructed from a vague phrase such as "compatible ranks"; instead, we use Haiman's precise theorem about these operators. The external theorem says the following. If $Q$ is a finite poset with $N$ elements and standard tableaux or linear-extension tableaux are identified with maximal chains of order ideals in $Q$, then Schützenberger promotion is conjugate to the composite \begin{align*} \varepsilon_N\circ\varepsilon_{N-1}:\mathcal C(Q)\to\mathcal C(Q). \end{align*} We verify the hypotheses in the present case. The poset is \begin{align*} P=\{1,\dots,a\}\times\{1,\dots,b\}, \end{align*} so it is finite and has $N=ab$ elements. The previous step constructed a bijection \begin{align*} \Phi:\operatorname{SYT}(\lambda)\to\mathcal C(P), \end{align*} so the required identification with maximal chains is available. Therefore Haiman's identity applies to $P$ and yields \begin{align*} \Phi\circ\operatorname{pr}\circ\Phi^{-1}=\varepsilon_N\circ\varepsilon_{N-1}. \end{align*} Unwinding this conjugacy gives, for every $T\in\operatorname{SYT}(\lambda)$, \begin{align*} \Phi(\operatorname{pr}(T))=(\varepsilon_N\circ\varepsilon_{N-1})(\Phi(T)). \end{align*} This is the exact bridge from tableaux to chains that will let the rectangular periodicity theorem act on $\Phi(T)$.[/guided]

[step:Apply Haiman's rectangular periodicity theorem]The only place where the shape being a rectangle is used is Haiman's rectangular dual-equivalence periodicity theorem. The theorem states: if $P=[a]\times[b]$ is the rectangle poset, $N=ab$, and \begin{align*} \varepsilon_m:\mathcal C(P)\to\mathcal C(P) \end{align*} are Haiman's partial evacuation operators, then \begin{align*} (\varepsilon_N\circ\varepsilon_{N-1})^N=\operatorname{id}_{\mathcal C(P)}. \end{align*} The hypotheses match the present setting because $P$ was defined as $\{1,\dots,a\}\times\{1,\dots,b\}$ with the product order, and $N=ab=|P|$. Hence, for every maximal chain $C\in\mathcal C(P)$, \begin{align*} (\varepsilon_N\circ\varepsilon_{N-1})^N(C)=C. \end{align*} External result used: Haiman's rectangular dual-equivalence periodicity theorem for maximal chains in a rectangular poset.[/step]

[guided]This step supplies the essential rectangular input. Haiman's rectangular dual-equivalence periodicity theorem says that for the rectangle poset \begin{align*} P=[a]\times[b] \end{align*} with $N=ab$ elements, the partial evacuation operators satisfy \begin{align*} (\varepsilon_N\circ\varepsilon_{N-1})^N=\operatorname{id}_{\mathcal C(P)}. \end{align*} We check that the theorem applies here. Our poset $P$ is exactly the rectangle \begin{align*} \{1,\dots,a\}\times\{1,\dots,b\} \end{align*} with product order, and the number of its elements is $N=ab$. The maps $\varepsilon_m$ are the same Haiman partial evacuation operators used in the promotion identity from the previous step. Therefore the theorem applies without any change of notation and gives, for each maximal chain $C\in\mathcal C(P)$, \begin{align*} (\varepsilon_N\circ\varepsilon_{N-1})^N(C)=C. \end{align*} This is not being derived from an informal cancellation sketch; it is the precise external theorem on which Haiman's proof of rectangular promotion periodicity rests.[/guided]

[step:Transport the chain identity back to tableaux]Let $T\in\operatorname{SYT}(\lambda)$, and set \begin{align*} C:=\Phi(T)\in\mathcal C(P). \end{align*} Using the promotion-partial-evacuation identity repeatedly gives \begin{align*} \Phi(\operatorname{pr}^N(T))=(\varepsilon_N\circ\varepsilon_{N-1})^N(C). \end{align*} By Haiman's rectangular periodicity theorem, \begin{align*} (\varepsilon_N\circ\varepsilon_{N-1})^N(C)=C. \end{align*} Thus \begin{align*} \Phi(\operatorname{pr}^N(T))=\Phi(T). \end{align*} Since $\Phi$ is a bijection, applying $\Phi^{-1}$ gives \begin{align*} \operatorname{pr}^N(T)=T. \end{align*} This holds for every $T\in\operatorname{SYT}(\lambda)$, so promotion has order dividing $N=ab$ on rectangular standard Young tableaux of shape $(b^a)$.[/step]

[guided]Choose an arbitrary tableau $T\in\operatorname{SYT}(\lambda)$ and define its associated maximal chain by \begin{align*} C:=\Phi(T)\in\mathcal C(P). \end{align*} The promotion-partial-evacuation identity gives one application of promotion as \begin{align*} \Phi(\operatorname{pr}(T))=(\varepsilon_N\circ\varepsilon_{N-1})(\Phi(T)). \end{align*} Applying the same identity repeatedly gives \begin{align*} \Phi(\operatorname{pr}^N(T))=(\varepsilon_N\circ\varepsilon_{N-1})^N(C). \end{align*} Now Haiman's rectangular periodicity theorem applies to the chain $C\in\mathcal C(P)$, so \begin{align*} (\varepsilon_N\circ\varepsilon_{N-1})^N(C)=C. \end{align*} Combining the two displayed identities yields \begin{align*} \Phi(\operatorname{pr}^N(T))=C=\Phi(T). \end{align*} Finally, the first step proved that \begin{align*} \Phi:\operatorname{SYT}(\lambda)\to\mathcal C(P) \end{align*} is a bijection. Therefore equality after applying $\Phi$ implies equality of the tableaux themselves, and hence \begin{align*} \operatorname{pr}^N(T)=T. \end{align*} Because $T$ was arbitrary, the identity holds for every $T\in\operatorname{SYT}(\lambda)$.[/guided]