[proofplan] We count semistandard tableaux by first forgetting the actual labels and retaining only their standardization. This partitions all semistandard tableaux of shape $\lambda$ with entries in $\{1,\dots,n\}$ into finitely many classes indexed by standard Young tableaux of shape $\lambda$. For each fixed standard tableau, the compatible choices of labels are integer points in a simplex cut out by finitely many weak and strict inequalities, so their number is a polynomial in $n$ of degree $r$ with leading coefficient $1/r!$. Summing over the $f_\lambda$ possible standardizations gives the leading term of $s_\lambda(1^n)$. [/proofplan]

[step:Standardize each semistandard tableau to a standard tableau] Let $Y(\lambda)$ denote the Young diagram of $\lambda$, viewed as the finite set of boxes \begin{align*} Y(\lambda) := \{(i,j) \in \mathbb{N}^2 : 1 \leq j \leq \lambda_i\}. \end{align*} A semistandard Young tableau of shape $\lambda$ with entries in $\{1,\dots,n\}$ is a map \begin{align*} S: Y(\lambda) &\to \{1,\dots,n\} \end{align*} such that $S(i,j) \leq S(i,j+1)$ whenever both boxes lie in $Y(\lambda)$ and $S(i,j) < S(i+1,j)$ whenever both boxes lie in $Y(\lambda)$. A standard Young tableau of shape $\lambda$ is a bijection \begin{align*} T: Y(\lambda) &\to \{1,\dots,r\} \end{align*} which is strictly increasing along rows and columns. Given a semistandard tableau $S$, order the boxes of $Y(\lambda)$ lexicographically by the triple \begin{align*} (S(i,j), -i, j). \end{align*} Thus boxes with equal entries are ordered from lower rows to higher rows, and within a fixed row from left to right. Assign the labels $1,\dots,r$ to the boxes in this order, and call the resulting bijection $\operatorname{std}(S):Y(\lambda)\to\{1,\dots,r\}$. We verify that $\operatorname{std}(S)$ is standard. If $(i,j)$ and $(i,j+1)$ are adjacent boxes in the same row, then $S(i,j)\leq S(i,j+1)$. If the inequality is strict, the first coordinate in the ordering puts $(i,j)$ before $(i,j+1)$; if equality holds, then the row indices are equal and $j<j+1$, so the third coordinate puts $(i,j)$ before $(i,j+1)$. Hence $\operatorname{std}(S)(i,j)<\operatorname{std}(S)(i,j+1)$. If $(i,j)$ and $(i+1,j)$ are adjacent boxes in the same column, then $S(i,j)<S(i+1,j)$ by semistandardness, so the first coordinate in the ordering gives $\operatorname{std}(S)(i,j)<\operatorname{std}(S)(i+1,j)$. Thus standardization defines a map from semistandard tableaux of shape $\lambda$ with entries in $\{1,\dots,n\}$ to standard Young tableaux of shape $\lambda$. [/step]

[step:Describe the semistandard tableaux with a fixed standardization]Fix a standard Young tableau \begin{align*} T: Y(\lambda) &\to \{1,\dots,r\}. \end{align*} For each $k\in\{1,\dots,r\}$, let $b_k\in Y(\lambda)$ be the unique box with $T(b_k)=k$. Define the descent set of $T$ by \begin{align*} D(T) := \{k\in\{1,\dots,r-1\}: b_{k+1} \text{ lies in a strictly lower row than } b_k\}. \end{align*} A semistandard tableau $S$ satisfies $\operatorname{std}(S)=T$ exactly when the sequence \begin{align*} a_T(S): \{1,\dots,r\} &\to \{1,\dots,n\}, \\ k &\mapsto S(b_k) \end{align*} satisfies \begin{align*} 1 \leq a_T(S)(1) \leq a_T(S)(2) \leq \cdots \leq a_T(S)(r) \leq n \end{align*} and \begin{align*} a_T(S)(k) < a_T(S)(k+1) \qquad \text{for every } k\in D(T). \end{align*} Indeed, the weak inequalities express that the entries of $S$ do not decrease in the standardization order. When $b_{k+1}$ lies in a lower row than $b_k$, equality would place $b_{k+1}$ before $b_k$ in the tie-breaking by the coordinate $-i$, contradicting $\operatorname{std}(S)=T$; hence strict inequality is required. Conversely, if such a sequence is given, define $S_T(a):Y(\lambda)\to\{1,\dots,n\}$ by \begin{align*} S_T(a)(b_k) := a(k). \end{align*} Because $T$ is increasing along rows, adjacent row boxes have labels $p<q$, so the weak chain gives $S_T(a)(b_p)\leq S_T(a)(b_q)$. Because $T$ is increasing down columns, adjacent column boxes have labels $p<q$ and the lower box lies in a strictly lower row than the upper box; since $p\in D(T)$ when $q=p+1$ and, more generally, some index $m\in\{p,\dots,q-1\}$ has $m\in D(T)$ along the label interval from the upper box to the lower box, the chain contains at least one strict inequality. Therefore $S_T(a)(b_p)<S_T(a)(b_q)$ for every adjacent column pair. Finally, the weak chain together with strictness at every $k\in D(T)$ says exactly that the order by the triples $(S_T(a)(i,j),-i,j)$ is the label order of $T$, so $\operatorname{std}(S_T(a))=T$.[/step]

[guided]Fix a standard tableau $T:Y(\lambda)\to\{1,\dots,r\}$. The point is to count all semistandard tableaux whose entries, after standardization, produce this particular $T$. Let $b_k$ be the box carrying the label $k$ in $T$. Any semistandard tableau $S$ with $\operatorname{std}(S)=T$ determines a sequence of actual entries \begin{align*} a_T(S): \{1,\dots,r\} &\to \{1,\dots,n\}, \\ k &\mapsto S(b_k). \end{align*} Since the labels $1,\dots,r$ were assigned in nondecreasing order of the semistandard entries, this sequence must satisfy \begin{align*} 1 \leq a_T(S)(1) \leq a_T(S)(2) \leq \cdots \leq a_T(S)(r) \leq n. \end{align*} There is one extra condition. Suppose $b_{k+1}$ lies in a strictly lower row than $b_k$. If $a_T(S)(k)=a_T(S)(k+1)$, then the standardization tie-breaker compares $-i$ before $j$, so the lower box $b_{k+1}$ is placed before the upper box $b_k$. That would reverse the desired order $k<k+1$ and contradict $\operatorname{std}(S)=T$. Thus equality is incompatible exactly at the descents \begin{align*} D(T) := \{k\in\{1,\dots,r-1\}: b_{k+1} \text{ lies in a strictly lower row than } b_k\}. \end{align*} Therefore every tableau in the fiber over $T$ satisfies \begin{align*} a_T(S)(k) < a_T(S)(k+1) \qquad \text{for every } k\in D(T). \end{align*} Conversely, if a sequence $a:\{1,\dots,r\}\to\{1,\dots,n\}$ satisfies the weak chain and the strict inequalities at $D(T)$, define a tableau \begin{align*} S_T(a):Y(\lambda)&\to\{1,\dots,n\},\\ b_k&\mapsto a(k). \end{align*} We first check the semistandard conditions. If two adjacent boxes lie in the same row and have labels $p<q$, then the weak chain gives $a(p)\leq a(q)$, so $S_T(a)$ is weakly increasing along rows. If an upper box and the adjacent lower box in the same column have labels $p<q$, then the row index starts at the upper row and ends at a strictly lower row as the boxes $b_p,b_{p+1},\dots,b_q$ are read in label order. Hence for some $m\in\{p,\dots,q-1\}$, the box $b_{m+1}$ lies in a strictly lower row than $b_m$, so $m\in D(T)$. The weak chain and the strict inequality at this $m$ give \begin{align*} a(p) \leq \cdots \leq a(m) < a(m+1) \leq \cdots \leq a(q), \end{align*} and therefore $a(p)<a(q)$. Thus $S_T(a)$ is strictly increasing down columns. Finally, the same inequalities recover the standardization. If $a(k)<a(k+1)$, the first coordinate in the triple $(S_T(a)(i,j),-i,j)$ puts $b_k$ before $b_{k+1}$. If $a(k)=a(k+1)$, then $k\notin D(T)$, so $b_{k+1}$ is not in a strictly lower row than $b_k$; the standardness of $T$ rules out a same-row reversal, and the tie-breaker by $-i$ and then $j$ puts $b_k$ before $b_{k+1}$. Hence consecutive labels appear in the order prescribed by $T$, so $\operatorname{std}(S_T(a))=T$.[/guided]

[step:Compute the leading term for each fixed standardization] For a fixed standard tableau $T$, let $d(T):=|D(T)|$. The number $N_T(n)$ of sequences \begin{align*} a:\{1,\dots,r\}\to\{1,\dots,n\} \end{align*} satisfying \begin{align*} 1 \leq a(1) \leq \cdots \leq a(r) \leq n \end{align*} and $a(k)<a(k+1)$ for all $k\in D(T)$ is a polynomial in $n$ of degree $r$ with leading coefficient $1/r!$. To see this, define \begin{align*} \varepsilon_k := \begin{cases} 1, & k\in D(T),\\ 0, & k\notin D(T), \end{cases} \qquad c_k := \varepsilon_1+\cdots+\varepsilon_{k-1} \end{align*} for $1\leq k\leq r$, with $c_1=0$. The map \begin{align*} a(k) \mapsto b(k):=a(k)-c_k \end{align*} is a bijection from the admissible sequences $a$ to weakly increasing sequences \begin{align*} 1 \leq b(1) \leq b(2) \leq \cdots \leq b(r) \leq n-d(T). \end{align*} The number of such weakly increasing sequences is the number of multisets of size $r$ chosen from $\{1,\dots,n-d(T)\}$, hence \begin{align*} N_T(n)=\binom{n-d(T)+r-1}{r}. \end{align*} As a polynomial in $n$, this has leading term \begin{align*} \binom{n-d(T)+r-1}{r} = \frac{1}{r!}n^r+O(n^{r-1}). \end{align*} Thus \begin{align*} \lim_{n\to\infty}\frac{N_T(n)}{n^r}=\frac{1}{r!}. \end{align*} [/step]

[step:Sum over all standard tableaux and take the limit] The standardization map partitions the semistandard tableaux of shape $\lambda$ with entries in $\{1,\dots,n\}$ by their standardization. Since $s_\lambda(1^n)$ is the number of such semistandard tableaux, the previous steps give \begin{align*} s_\lambda(1^n) = \sum_T N_T(n), \end{align*} where the sum ranges over all standard Young tableaux $T$ of shape $\lambda$. There are exactly $f_\lambda$ such tableaux. Therefore \begin{align*} \frac{s_\lambda(1^n)}{n^r} = \sum_T \frac{N_T(n)}{n^r}. \end{align*} Taking $n\to\infty$ and using the fixed finite number $f_\lambda$ of summands gives \begin{align*} \lim_{n\to\infty}\frac{s_\lambda(1^n)}{n^r} = \sum_T \frac{1}{r!} = \frac{f_\lambda}{r!}. \end{align*} Multiplying both sides by $r!$ yields \begin{align*} f_\lambda = r!\lim_{n\to\infty}\frac{s_\lambda(1^n)}{n^r}. \end{align*} This is the claimed identity. [/step]