[proofplan] We compare two asymptotic evaluations of the Schur specialization $s_\lambda(1^n)$ as $n \to \infty$. The [hook-content formula](/theorems/5211) expresses this specialization as an explicit product whose leading term is determined by the hook lengths. On the other hand, the standard-tableaux specialization limit identifies the leading coefficient of $s_\lambda(1^n)$ with \begin{align*} \frac{f^\lambda}{r!}. \end{align*} Equating the two leading coefficients gives the hook-length formula. [/proofplan] [step:Introduce the hook and content data attached to the Young diagram] Let $\ell := \ell(\lambda)$ denote the number of nonzero parts of the partition $\lambda$. Write the Young diagram of $\lambda$ as \begin{align*} [\lambda] := \{(i,j) \in \mathbb{N}^2 : 1 \leq i \leq \ell,\ 1 \leq j \leq \lambda_i\}. \end{align*} Thus $|[\lambda]| = r$. For each box $u = (i,j) \in [\lambda]$, define its content $c(u) \in \mathbb{Z}$ by \begin{align*} c(u) := j - i. \end{align*} For each box $u = (i,j) \in [\lambda]$, define its hook length $h(u) \in \mathbb{N}$ by \begin{align*} h(u) := \#\{(i,k) \in [\lambda] : k \geq j\} + \#\{(k,j) \in [\lambda] : k > i\}. \end{align*} Define the hook product $H_\lambda \in \mathbb{N}$ by \begin{align*} H_\lambda := \prod_{u \in [\lambda]} h(u). \end{align*} Since every hook length $h(u)$ is a positive integer, $H_\lambda > 0$. For each $n \in \mathbb{N}$, let $s_\lambda(1^n)$ denote the specialization of the Schur polynomial $s_\lambda$ obtained by setting the first $n$ variables equal to $1$ and all remaining variables equal to $0$. [/step] [step:Extract the leading coefficient from the hook-content formula] We use the hook-content formula for Schur polynomial specializations as an established prerequisite: for every partition $\mu$ and every $n \in \mathbb{N}$, the specialization $s_\mu(1^n)$ is given by the product over the boxes of $[\mu]$ with numerator $n$ plus the content and denominator the hook length. The present $\lambda$ is a partition, and $n \in \mathbb{N}$, so this formula applies and gives \begin{align*} s_\lambda(1^n) = \prod_{u \in [\lambda]} \frac{n + c(u)}{h(u)} = \frac{1}{H_\lambda}\prod_{u \in [\lambda]}(n + c(u)). \end{align*} Because $[\lambda]$ has exactly $r$ boxes, the product $\prod_{u \in [\lambda]}(n + c(u))$ is a monic polynomial in $n$ of degree $r$. Therefore \begin{align*} \lim_{n \to \infty} \frac{s_\lambda(1^n)}{n^r} = \frac{1}{H_\lambda} = \frac{1}{\prod_{u \in [\lambda]} h(u)}. \end{align*} [guided] We briefly repeat the notation needed inside this step. Let $\ell := \ell(\lambda)$ denote the number of nonzero parts of $\lambda$. The Young diagram is \begin{align*} [\lambda] := \{(i,j) \in \mathbb{N}^2 : 1 \leq i \leq \ell,\ 1 \leq j \leq \lambda_i\}, \end{align*} and for a box $u = (i,j) \in [\lambda]$ the content is $c(u) := j-i$. The hook length $h(u) \in \mathbb{N}$ is \begin{align*} h(u) := \#\{(i,k) \in [\lambda] : k \geq j\} + \#\{(k,j) \in [\lambda] : k > i\}. \end{align*} We use the hook-content formula for Schur polynomial specializations as an established prerequisite. It applies to every partition and every $n \in \mathbb{N}$; since the present $\lambda$ is a partition and $n \in \mathbb{N}$, it gives \begin{align*} s_\lambda(1^n) = \prod_{u \in [\lambda]} \frac{n + c(u)}{h(u)}. \end{align*} For this proof, we only need the leading term of this expression as a polynomial in $n$. The denominator is independent of $n$, so we collect it into the hook product \begin{align*} H_\lambda = \prod_{u \in [\lambda]} h(u). \end{align*} This gives \begin{align*} s_\lambda(1^n) = \frac{1}{H_\lambda}\prod_{u \in [\lambda]}(n + c(u)). \end{align*} There are exactly $r$ boxes in $[\lambda]$, because $\lambda$ is a partition of $r$. Each factor $n + c(u)$ is a monic degree-one polynomial in $n$. Hence their product is a monic degree-$r$ polynomial in $n$. Dividing by $n^r$ and taking $n \to \infty$ retains only this leading coefficient: \begin{align*} \lim_{n \to \infty} \frac{1}{n^r}\prod_{u \in [\lambda]}(n + c(u)) = 1. \end{align*} Multiplying by the constant factor $1/H_\lambda$, we obtain \begin{align*} \lim_{n \to \infty} \frac{s_\lambda(1^n)}{n^r} = \frac{1}{H_\lambda} = \frac{1}{\prod_{u \in [\lambda]} h(u)}. \end{align*} [/guided] [/step] [step:Identify the same leading coefficient by standard tableaux] We use the standard-tableaux specialization asymptotic for Schur polynomials as an established prerequisite: for every fixed partition $\mu$ of size $m$, the leading coefficient of $s_\mu(1^n)$ as a polynomial in $n$ is the number of standard Young tableaux of shape $\mu$ divided by $m!$. The present $\lambda$ is a fixed partition of size $r$, so this asymptotic applies and gives \begin{align*} \lim_{n \to \infty} \frac{s_\lambda(1^n)}{n^r} = \frac{f^\lambda}{r!}, \end{align*} where $f^\lambda$ denotes the number of standard Young tableaux of shape $\lambda$. [/step] [step:Equate the two limits and solve for $f^\lambda$] The preceding two steps compute the same limit. Hence \begin{align*} \frac{f^\lambda}{r!} = \frac{1}{\prod_{u \in [\lambda]} h(u)}. \end{align*} Multiplying both sides by $r!$ gives \begin{align*} f^\lambda = \frac{r!}{\prod_{u \in [\lambda]} h(u)}. \end{align*} Since $[\lambda]$ is the Young diagram of $\lambda$, this is precisely \begin{align*} f^\lambda = \frac{r!}{\prod_{u \in \lambda} h(u)}. \end{align*} This proves the Frame-Robinson-Thrall hook-length formula. [/step]