[proofplan] We realize a tableau as its row reading word, so the tableau operators are the restrictions of the standard type $A_{n-1}$ word-crystal operators. The [tensor product](/page/Tensor%20Product) rule gives precisely the signature cancellation rule on this reading word, and semistandardness is preserved whenever an operator is defined. The canonical tableau whose $r$th row is filled with $r$ is the unique highest-weight vertex of weight $\lambda$, and every semistandard tableau raises to it by repeated $e_i$ operators. Kashiwara uniqueness then identifies this connected normal highest-weight crystal with the representation-theoretic crystal of the irreducible polynomial $GL_n(\mathbb C)$-module of highest weight $\lambda$. [/proofplan]

[step:Embed tableaux into the word crystal using the row reading word] Let $A_n=\{1,\dots,n\}$ with its usual order. Let $0$ be the formal null crystal element declared in the theorem statement. Let $B=A_n$ be the single-letter type $A_{n-1}$ crystal, with weight map $\operatorname{wt}:B\to \mathbb Z^n$ defined by $\operatorname{wt}(a)=\varepsilon_a$, where $(\varepsilon_1,\dots,\varepsilon_n)$ is the standard basis of $\mathbb Z^n$. For a word $w=w_1\cdots w_m\in A_n^m$, regard $w$ as the tensor element \begin{align*} w_1\otimes \cdots \otimes w_m \in B^{\otimes m}. \end{align*} Let $|\lambda|=\lambda_1+\cdots+\lambda_n$. Define \begin{align*} \operatorname{rw}: \operatorname{SSYT}_n(\lambda) &\to A_n^{|\lambda|} \end{align*} to be the map sending a tableau to the word obtained by reading rows from bottom to top and, inside each row, from left to right. This map is injective, because the shape $\lambda$ is fixed and the row lengths determine which consecutive subwords of $\operatorname{rw}(T)$ belong to which rows. For each $i\in\{1,\dots,n-1\}$, every single-letter $i$-string in $B$ has length at most $1$, so the finite-string hypothesis of [citetheorem:8459] is satisfied. Applying [citetheorem:8459] inductively to the tensor product $B^{\otimes |\lambda|}$, with tensor factors read from left to right, shows that the operators on $B^{\otimes |\lambda|}$ are computed by the usual signature rule on the word: retain only letters $i$ and $i+1$, write $+$ for $i$ and $-$ for $i+1$, cancel adjacent pairs $-+$ repeatedly, and then apply $e_i$ to the letter corresponding to the leftmost uncancelled $-$, or apply $f_i$ to the letter corresponding to the rightmost uncancelled $+$, when such a sign exists. Therefore the tableau operators in the statement are precisely the restrictions of the word-crystal operators through the embedding $\operatorname{rw}$. [/step]

[step:Check that the highest tableau is killed by every raising operator]Define the tableau $H_\lambda\in \operatorname{SSYT}_n(\lambda)$ by filling every box in row $r$ with the entry $r$, for $1\le r\le n$. This is semistandard: each row is constant, hence weakly increasing, and entries strictly increase down columns because row $r+1$ contains $r+1>r$. The weight of $H_\lambda$ is \begin{align*} \operatorname{wt}(H_\lambda)=\lambda_1\varepsilon_1+\cdots+\lambda_n\varepsilon_n, \end{align*} which we identify with the partition $\lambda$ in the polynomial $GL_n$ weight lattice. Fix $i\in\{1,\dots,n-1\}$. In the row reading word of $H_\lambda$, all entries $i+1$ occur before all entries $i$, because row $i+1$ is read before row $i$. Thus the $i$-signature consists of $\lambda_{i+1}$ signs $-$ followed by $\lambda_i$ signs $+$. Since $\lambda_i\ge \lambda_{i+1}$, repeated cancellation of adjacent pairs $-+$ removes every $-$ and leaves $\lambda_i-\lambda_{i+1}$ uncancelled $+$ signs. Hence there is no uncancelled $-$, so $e_i(H_\lambda)=0$. Since this holds for every $i$, the tableau $H_\lambda$ is a highest-weight vertex of weight $\lambda$.[/step]

[guided]The point of choosing $H_\lambda$ is that it is the tableau whose entries are as small as the shape allows: row $1$ is filled by $1$, row $2$ is filled by $2$, and so on. First we verify that this is a legitimate semistandard tableau. Rows are weakly increasing because every row is constant. Columns are strictly increasing because moving from row $r$ to row $r+1$ changes the entry from $r$ to $r+1$, and $r<r+1$. Now compute its weight. The weight records how many times each letter occurs. Since row $r$ has $\lambda_r$ boxes and every entry in that row is $r$, the content vector is \begin{align*} \operatorname{wt}(H_\lambda)=\lambda_1\varepsilon_1+\cdots+\lambda_n\varepsilon_n. \end{align*} This is the usual identification of a partition with at most $n$ parts as a dominant polynomial weight for $GL_n$. It remains to check the highest-weight condition. Fix $i\in\{1,\dots,n-1\}$. Under the chosen row reading convention, lower rows are read first. Therefore every entry $i+1$ in row $i+1$ appears before every entry $i$ in row $i$. When we form the $i$-signature, the retained subword has $\lambda_{i+1}$ copies of $i+1$ followed by $\lambda_i$ copies of $i$, so the signature has the form \begin{align*} \underbrace{-\cdots -}_{\lambda_{i+1}\text{ signs}}\underbrace{+\cdots +}_{\lambda_i\text{ signs}}. \end{align*} The cancellation rule deletes adjacent pairs $-+$. Because $\lambda_i\ge \lambda_{i+1}$, every $-$ is paired with a later $+$ and is cancelled. Thus no uncancelled $-$ remains. By definition of the word-crystal operator, $e_i$ is defined exactly when an uncancelled $-$ remains, so $e_i(H_\lambda)=0$. This argument works for every simple root index $i$, hence $H_\lambda$ is highest weight.[/guided]

[step:Restrict the word-crystal operators to semistandard tableaux] Let $i\in\{1,\dots,n-1\}$. The tableau operator $e_i$ changes, when defined, a single entry $i+1$ to $i$ at the box corresponding through $\operatorname{rw}$ to the selected uncancelled $-$ in the reduced signature. Similarly, $f_i$ changes, when defined, a single entry $i$ to $i+1$ at the box corresponding to the selected uncancelled $+$. The hypotheses of [citetheorem:8456] are exactly that $T\in\operatorname{SSYT}_n(\lambda)$ and that one of the tableau signature operators $e_i(T)$ or $f_i(T)$ is defined. Hence [citetheorem:8456] applies here and gives that the resulting filling is again an element of $\operatorname{SSYT}_n(\lambda)$. Thus $\operatorname{SSYT}_n(\lambda)\sqcup\{0\}$ is closed under all operators $e_i$ and $f_i$. If $|\lambda|=0$, the inverse property is immediate because the only tableau is the empty tableau and every operator has value $0$. If $|\lambda|>0$, then the word length $|\lambda|$ is a positive integer, and for every relevant simple root index $i$ we have $n\ge 2$; therefore the hypotheses of [citetheorem:8455] are satisfied for the word set $\{1,\dots,n\}^{|\lambda|}$. Because the word-crystal operators satisfy the inverse property on words by [citetheorem:8455], and because $\operatorname{rw}$ is injective, the restricted tableau operators also satisfy \begin{align*} e_i(T')=T \iff f_i(T)=T' \end{align*} for all $T,T'\in \operatorname{SSYT}_n(\lambda)$ and all $i\in\{1,\dots,n-1\}$. Hence the signature operators give $\operatorname{SSYT}_n(\lambda)$ the structure of a type $A_{n-1}$ crystal. [/step]

[step:Show that every semistandard tableau raises to the highest tableau]Let $T\in\operatorname{SSYT}_n(\lambda)$. If $|\lambda|=0$, then $\operatorname{SSYT}_n(\lambda)$ consists of the empty tableau, which is $H_\lambda$, so the assertion holds. Assume now that $|\lambda|>0$. Since $\operatorname{SSYT}_n(\lambda)$ is finite, any sequence of successive applications of raising operators $e_i$ must terminate. Choose a tableau $T^\sharp\in\operatorname{SSYT}_n(\lambda)$ obtained from $T$ by applying a finite sequence of operators $e_i$ such that \begin{align*} e_i(T^\sharp)=0 \end{align*} for every $i\in\{1,\dots,n-1\}$. We prove that $T^\sharp=H_\lambda$. Let $w=\operatorname{rw}(T^\sharp)$. Since all $e_i(T^\sharp)$ vanish, the reduced $i$-signature of $w$ has no uncancelled $-$ for every $i\in\{1,\dots,n-1\}$. With the present convention, this is equivalent to the following suffix ballot condition: for every suffix $u$ of $w$ and every $i\in\{1,\dots,n-1\}$, the number of letters $i$ in $u$ is at least the number of letters $i+1$ in $u$. Indeed, after cancelling adjacent pairs $-+$, an uncancelled $-$ remains exactly when some suffix of the sign string contains more $-$ signs than $+$ signs. Now use the row reading order. The last letters of $w$ are the entries of the top row of $T^\sharp$, read from left to right. Applying the suffix condition to the one-letter suffix consisting of the rightmost top-row entry shows that this entry must be $1$, since a one-letter suffix containing $a>1$ has more letters $a$ than letters $a-1$. Since the top row is weakly increasing, every entry in the top row is at most this rightmost entry and hence equals $1$. Remove the top $r-1$ rows from consideration, and suppose inductively that rows $1,\dots,r-1$ have been shown to be filled with $1,\dots,r-1$, respectively. Strict increase down columns forces every entry in row $r$ to be at least $r$. If some entry of row $r$ were $a>r$, choose the leftmost such entry and let $u$ be the suffix of $w$ beginning at that box. Since rows are weakly increasing, every later entry in row $r$ is at least $a$. Since the rows above row $r$ have already been identified, they contain only letters in $\{1,\dots,r-1\}$. Thus $u$ contains at least one letter $a$ and contains no letter $a-1$, contradicting the suffix ballot condition for $i=a-1$. Therefore every entry in row $r$ is $r$. By induction, each row $r$ of $T^\sharp$ is filled by $r$, so $T^\sharp=H_\lambda$. Consequently every $T\in\operatorname{SSYT}_n(\lambda)$ lies in the connected component generated from $H_\lambda$ by the lowering operators $f_i$, and $H_\lambda$ is the unique highest-weight vertex.[/step]

[guided]The delicate point is that the ballot condition must match the chosen reading convention. We read rows from bottom to top and cancel adjacent pairs $-+$, where $-$ denotes $i+1$ and $+$ denotes $i$. Under this convention, highest weight is a suffix condition, not a prefix condition: for every suffix $u$ of the row reading word and every $i\in\{1,\dots,n-1\}$, the suffix $u$ contains at least as many letters $i$ as letters $i+1$. This follows directly from the cancellation rule, because an uncancelled $-$ remains precisely when some suffix of the reduced sign string has more $-$ signs than $+$ signs. Let $T^\sharp$ be a terminal tableau obtained from $T$ by repeatedly applying raising operators. Terminal means \begin{align*} e_i(T^\sharp)=0 \end{align*} for every $i\in\{1,\dots,n-1\}$, so the suffix ballot condition applies to $w=\operatorname{rw}(T^\sharp)$. The last letters of $w$ are the entries of the top row. The one-letter suffix consisting of the rightmost entry of the top row cannot be $a>1$, because then it would contain one letter $a$ and no letter $a-1$, violating the suffix condition for $i=a-1$. Hence the rightmost top-row entry is $1$, and weak row increase forces the whole top row to be $1$. Now proceed row by row. Suppose rows $1,\dots,r-1$ have already been shown to contain exactly their row indices. Strict column increase forces every entry in row $r$ to be at least $r$. If the first entry in row $r$ larger than $r$ were $a>r$, consider the suffix beginning at that box. Weak increase along row $r$ makes every later entry in row $r$ at least $a$, while the rows above row $r$ have already been shown to contain only the letters $1,\dots,r-1$. Hence this suffix contains at least one letter $a$ and contains no letter $a-1$. This contradicts the suffix ballot condition for $i=a-1$. Therefore no entry in row $r$ exceeds $r$, and strict columns already gave no entry below $r$, so row $r$ is filled entirely with $r$. Induction gives $T^\sharp=H_\lambda$.[/guided]

[step:Identify the connected tableau crystal with the irreducible highest-weight crystal] The preceding steps show that $\operatorname{SSYT}_n(\lambda)$ is a connected type $A_{n-1}$ crystal with unique highest-weight vertex $H_\lambda$ of weight $\lambda$. Since $\operatorname{SSYT}_n(\lambda)$ is closed under the restricted word-crystal operators, it is a subcrystal of the tensor product $B^{\otimes |\lambda|}$, and the connectedness just proved shows that it is a connected component of the subcrystal generated by $H_\lambda$. Each single-letter crystal is normal, and the tensor product rule preserves normality, so $B^{\otimes |\lambda|}$ is normal. For every $i\in\{1,\dots,n-1\}$, the full $i$-string through a tableau remains inside $\operatorname{SSYT}_n(\lambda)$ because closure holds for both $e_i$ and $f_i$; hence the finite $i$-string data in the tableau crystal agrees with the corresponding $i$-string data in the normal word crystal. Therefore the tableau crystal is normal. The theorem statement defines $B_n(\lambda)$ as the connected normal type $A_{n-1}$ highest-weight crystal with unique highest-weight vertex of dominant polynomial weight $\lambda$. The hypotheses of [citetheorem:8473] are now verified: the tableau crystal is connected, normal, and has unique highest-weight vertex $H_\lambda$ of weight $\lambda$. By [citetheorem:8473], it is uniquely isomorphic to $B_n(\lambda)$. Therefore there is an isomorphism of crystals \begin{align*} \operatorname{SSYT}_n(\lambda)\cong B_n(\lambda). \end{align*} For polynomial type $A$, the dominant weight $\lambda=(\lambda_1,\dots,\lambda_n)$ indexes the irreducible polynomial representation of $GL_n(\mathbb C)$ of highest weight $\lambda$, and its crystal is precisely $B_n(\lambda)$. Hence the semistandard tableaux of shape $\lambda$ with entries in $\{1,\dots,n\}$, equipped with the stated signature operators and content weights, realize the highest-weight crystal of that irreducible representation. This proves the theorem. [/step]