[proofplan] We begin from the principal specialization formula for the Schur polynomial $s_\lambda(1,q,\dots,q^{n-1})$. The main work is a row-by-row reindexing of the Weyl-type product into the hook-content product, proved by an explicit multiset identity for the exponents. Once this $q$-identity is established, taking the limit $q \to 1$ turns each factor $(1-q^A)/(1-q^B)$ into $A/B$. Finally, evaluating the tableau generating function at $x_1=\cdots=x_n=1$ identifies $s_\lambda(1^n)$ with the desired number of semistandard tableaux. [/proofplan]

[step:Start from the principal specialization formula for Schur polynomials] Let $r := \ell(\lambda)$ denote the number of nonzero parts of $\lambda$, and extend the partition by setting $\lambda_j := 0$ for every integer $j > r$. Let $q$ be an indeterminate over $\mathbb{Q}$, so all expressions below are interpreted in the rational function field $\mathbb{Q}(q)$. Define \begin{align*} n(\lambda) := \sum_{i=1}^{n} (i-1)\lambda_i. \end{align*} We use the principal specialization formula for [Schur polynomials](/page/Schur%20Polynomial) as an external prerequisite: for every partition with at most $n$ parts, its principal specialization is the rational function identity \begin{align*} s_\lambda(1,q,\dots,q^{n-1}) = q^{n(\lambda)} \prod_{1 \leq i < j \leq n} \frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}}. \end{align*} [/step]

[step:Rewrite each row factor as a hook-content product]For each fixed row index $i \in \{1,\dots,n\}$, define the row factor \begin{align*} P_i(q) := \prod_{j=i+1}^{n} \frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}}, \end{align*} where the empty product is $1$ when $i=n$. We prove that \begin{align*} P_i(q) = \prod_{a=1}^{\lambda_i} \frac{1-q^{n+a-i}}{1-q^{\lambda_i-a+\lambda'_a-i+1}}. \end{align*} If $\lambda_i=0$, then $\lambda_j=0$ for every $j>i$, so each factor in $P_i(q)$ is \begin{align*} \frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}} = \frac{1-q^{j-i}}{1-q^{j-i}} = 1. \end{align*} Thus $P_i(q)=1$, and the right-hand side is the empty product $1$ because there are no cells in row $i$. Assume now that $\lambda_i>0$. Define integers \begin{align*} A_k := \lambda_i-\lambda_k+k-i \end{align*} for $k \in \{i,\dots,n+1\}$, where $\lambda_{n+1}:=0$. Then $A_i=0$ and \begin{align*} A_{n+1}=\lambda_i+n+1-i. \end{align*} Moreover, since $\lambda$ is weakly decreasing, \begin{align*} A_{k+1}-A_k=\lambda_k-\lambda_{k+1}+1\geq 1 \end{align*} for every $k\in\{i,\dots,n\}$. Thus the sequence $A_i,A_{i+1},\dots,A_{n+1}$ is strictly increasing, so the intervals used below are ordered correctly. The numerator exponents in $P_i(q)$ are precisely \begin{align*} A_{i+1},A_{i+2},\dots,A_n, \end{align*} while the denominator exponents in $P_i(q)$ are \begin{align*} 1,2,\dots,n-i. \end{align*} Now consider the hook exponents in row $i$. For a fixed $k \in \{i,\dots,n\}$, the columns $a$ satisfying \begin{align*} \lambda_{k+1}<a\leq \lambda_k \end{align*} are exactly those for which $\lambda'_a=k$. For these columns, \begin{align*} \lambda_i-a+\lambda'_a-i+1 = \lambda_i-a+k-i+1. \end{align*} As $a$ runs from $\lambda_{k+1}+1$ to $\lambda_k$, these exponents run through the interval \begin{align*} A_k+1,\ A_k+2,\ \dots,\ A_{k+1}-1. \end{align*} Therefore the multiset consisting of the numerator exponents of $P_i(q)$ together with the hook exponents in row $i$ is \begin{align*} \{1,2,\dots,\lambda_i+n-i\}. \end{align*} The same multiset is obtained by combining the denominator exponents of $P_i(q)$ with the content exponents \begin{align*} n+a-i,\qquad 1\leq a\leq \lambda_i, \end{align*} because \begin{align*} \{1,\dots,n-i\}\cup\{n+1-i,\dots,n+\lambda_i-i\} = \{1,\dots,\lambda_i+n-i\}. \end{align*} Thus the products of the corresponding factors $1-q^m$ agree: \begin{align*} \prod_{j=i+1}^{n}\bigl(1-q^{\lambda_i-\lambda_j+j-i}\bigr) \prod_{a=1}^{\lambda_i} \bigl(1-q^{\lambda_i-a+\lambda'_a-i+1}\bigr) = \prod_{j=i+1}^{n}\bigl(1-q^{j-i}\bigr) \prod_{a=1}^{\lambda_i} \bigl(1-q^{n+a-i}\bigr). \end{align*} Dividing by the two denominator products gives the claimed row identity.[/step]

[guided]We want to explain why the quotient over pairs $i<j$ reorganizes into hook and content factors. Fix a row index $i \in \{1,\dots,n\}$ and isolate all factors involving this row: \begin{align*} P_i(q) := \prod_{j=i+1}^{n} \frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}}. \end{align*} The goal is to prove \begin{align*} P_i(q) = \prod_{a=1}^{\lambda_i} \frac{1-q^{n+a-i}}{1-q^{\lambda_i-a+\lambda'_a-i+1}}. \end{align*} Here $\lambda'_a:=\#\{k\in\{1,\dots,n\}:\lambda_k\geq a\}$ denotes the length of column $a$ in the conjugate partition. The exponent $n+a-i$ is $n+c(i,a)$, while the exponent $\lambda_i-a+\lambda'_a-i+1$ is the hook length $h(i,a)$. If $\lambda_i=0$, then every lower row also has length $0$, so the numerator and denominator products defining $P_i(q)$ cancel term by term: \begin{align*} \lambda_i-\lambda_j+j-i=j-i. \end{align*} The right-hand side is also an empty product because there are no cells in row $i$. Hence the identity holds in this case. Assume $\lambda_i>0$. Define \begin{align*} A_k := \lambda_i-\lambda_k+k-i \end{align*} for $k \in \{i,\dots,n+1\}$, with $\lambda_{n+1}:=0$. These numbers mark the exponents appearing when the lower row index reaches $k$. In particular, \begin{align*} A_i=0, \qquad A_{n+1}=\lambda_i+n+1-i. \end{align*} Since $\lambda$ is weakly decreasing, for every $k\in\{i,\dots,n\}$ we have \begin{align*} A_{k+1}-A_k=\lambda_k-\lambda_{k+1}+1\geq 1. \end{align*} Thus the sequence $A_i,A_{i+1},\dots,A_{n+1}$ is strictly increasing, and the gaps between consecutive $A_k$ are genuine ordered intervals. The numerator exponents in $P_i(q)$ are exactly \begin{align*} A_{i+1},A_{i+2},\dots,A_n, \end{align*} because \begin{align*} \lambda_i-\lambda_j+j-i=A_j. \end{align*} The denominator exponents in $P_i(q)$ are simply \begin{align*} 1,2,\dots,n-i. \end{align*} Now we identify the hook exponents row by row. Fix $k \in \{i,\dots,n\}$. The condition \begin{align*} \lambda_{k+1}<a\leq \lambda_k \end{align*} means exactly that column $a$ reaches row $k$ but not row $k+1$. Therefore $\lambda'_a=k$. For such a column $a$, the hook exponent in row $i$ is \begin{align*} h(i,a) = \lambda_i-a+\lambda'_a-i+1 = \lambda_i-a+k-i+1. \end{align*} As $a$ increases from $\lambda_{k+1}+1$ to $\lambda_k$, the value of this expression decreases through the integers \begin{align*} A_{k+1}-1,\ A_{k+1}-2,\ \dots,\ A_k+1. \end{align*} Equivalently, the hook exponents contributed by these columns are the interval \begin{align*} A_k+1,\ A_k+2,\ \dots,\ A_{k+1}-1. \end{align*} Thus, for each $k$, the hook exponents fill exactly the gaps between $A_k$ and $A_{k+1}$. The numerator exponents $A_{i+1},\dots,A_n$ supply the internal endpoints of these gaps. Since $A_i=0$ and $A_{n+1}=\lambda_i+n+1-i$, the multiset formed by the numerator exponents of $P_i(q)$ together with all hook exponents in row $i$ is \begin{align*} \{1,2,\dots,\lambda_i+n-i\}. \end{align*} The denominator exponents of $P_i(q)$ give the initial block \begin{align*} 1,2,\dots,n-i, \end{align*} while the content exponents in row $i$ give the final block \begin{align*} n+1-i,\ n+2-i,\dots,n+\lambda_i-i. \end{align*} Together these also form \begin{align*} \{1,2,\dots,\lambda_i+n-i\}. \end{align*} Therefore the two products over factors $1-q^m$ are equal: \begin{align*} \prod_{j=i+1}^{n}\bigl(1-q^{\lambda_i-\lambda_j+j-i}\bigr) \prod_{a=1}^{\lambda_i} \bigl(1-q^{h(i,a)}\bigr) = \prod_{j=i+1}^{n}\bigl(1-q^{j-i}\bigr) \prod_{a=1}^{\lambda_i} \bigl(1-q^{n+a-i}\bigr). \end{align*} Dividing by the product of the denominator factors gives \begin{align*} P_i(q) = \prod_{a=1}^{\lambda_i} \frac{1-q^{n+a-i}}{1-q^{h(i,a)}}. \end{align*} This is the desired row-wise hook-content identity.[/guided]

[step:Multiply the row identities to obtain the principal hook-content specialization] Multiplying the row identity over $i=1,\dots,n$ gives \begin{align*} \prod_{1 \leq i < j \leq n} \frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}} = \prod_{i=1}^{n} \prod_{a=1}^{\lambda_i} \frac{1-q^{n+a-i}}{1-q^{\lambda_i-a+\lambda'_a-i+1}}. \end{align*} Since the cells of $\lambda$ are exactly the pairs $u=(i,a)$ with $1\leq i\leq n$ and $1\leq a\leq \lambda_i$, and since \begin{align*} c(i,a)&=a-i,\\ h(i,a)&=\lambda_i-a+\lambda'_a-i+1, \end{align*} this becomes \begin{align*} \prod_{1 \leq i < j \leq n} \frac{1-q^{\lambda_i-\lambda_j+j-i}}{1-q^{j-i}} = \prod_{u\in\lambda} \frac{1-q^{n+c(u)}}{1-q^{h(u)}}. \end{align*} Substituting this into the principal specialization formula yields \begin{align*} s_\lambda(1,q,\dots,q^{n-1}) = q^{n(\lambda)} \prod_{u\in\lambda} \frac{1-q^{n+c(u)}}{1-q^{h(u)}}. \end{align*} [/step]

[step:Take the limit as $q$ tends to $1$] We now regard $q$ as a real variable in a punctured neighbourhood of $1$, where the rational function identity obtained above is valid after evaluation. For every cell $u=(i,a)\in\lambda$, the integers $n+c(u)=n+a-i$ and $h(u)$ are positive. Indeed $1\leq i\leq r\leq n$ and $a\geq 1$, so $n+a-i\geq 1$, and $h(u)$ is the number of cells in the hook through $u$, hence $h(u)\geq 1$. For positive integers $A$ and $B$, \begin{align*} \frac{1-q^A}{1-q^B} = \frac{1+q+\cdots+q^{A-1}}{1+q+\cdots+q^{B-1}} \end{align*} for $q\neq 1$, and therefore \begin{align*} \lim_{q\to 1}\frac{1-q^A}{1-q^B} = \frac{A}{B}. \end{align*} Also, \begin{align*} \lim_{q\to 1}q^{n(\lambda)}=1. \end{align*} Taking the limit $q\to 1$ in the hook-content specialization gives \begin{align*} s_\lambda(1^n) = \prod_{u\in\lambda} \frac{n+c(u)}{h(u)}. \end{align*} [/step]

[step:Identify the specialization with the number of semistandard tableaux] By the tableau definition of the [Schur polynomial](/page/Schur%20Polynomial), \begin{align*} s_\lambda(x_1,\dots,x_n) = \sum_T \prod_{u\in\lambda} x_{T(u)}, \end{align*} where the sum is over all [semistandard Young tableaux](/page/Semistandard%20Young%20Tableau) $T:\lambda\to\{1,\dots,n\}$ of shape $\lambda$. Setting $x_1=\cdots=x_n=1$ makes each monomial equal to $1$, so \begin{align*} s_\lambda(1^n) = \#\{T:\lambda\to\{1,\dots,n\}: T \text{ is semistandard}\}. \end{align*} Combining this identification with the product formula obtained above proves that the number of semistandard Young tableaux of shape $\lambda$ with entries in $\{1,\dots,n\}$ is \begin{align*} \prod_{u\in\lambda}\frac{n+c(u)}{h(u)}. \end{align*} [/step]