[proofplan] The proof is a restriction of the RSK bijection to one fixed shape. The RSK correspondence identifies each permutation $w\in S_n$ with an ordered pair $(P(w),Q(w))$ of standard Young tableaux of the same shape. The condition $\operatorname{sh}(w)=\lambda$ is exactly the condition that both tableaux have shape $\lambda$, so the desired fibre is in bijection with ordered pairs of standard Young tableaux of shape $\lambda$. Counting the first and second tableau independently gives $(f_\lambda)^2$. [/proofplan] [step:Restrict the RSK bijection to permutations of shape $\lambda$] Let \begin{align*} A_\lambda:=\{w\in S_n:\operatorname{sh}(w)=\lambda\} \end{align*} be the set of permutations whose RSK shape is $\lambda$. Let \begin{align*} \operatorname{SYT}(\lambda):=\{T:T \text{ is a standard Young tableau of shape } \lambda\} \end{align*} be the set of standard Young tableaux of shape $\lambda$, so that $f_\lambda=|\operatorname{SYT}(\lambda)|$ by definition. Let \begin{align*} \mathcal{C}_n:=\{(P,Q):\text{there exists }\mu\vdash n\text{ such that }P,Q\in\operatorname{SYT}(\mu)\} \end{align*} denote the set of ordered pairs of standard Young tableaux of the same shape and total size $n$. By [citetheorem:8435], the map $\operatorname{RSK}:S_n\to \mathcal{C}_n$ given by $w\mapsto (P(w),Q(w))$ is a bijection. Since $\operatorname{sh}(w)$ is the common shape of $P(w)$ and $Q(w)$, the restriction of $\operatorname{RSK}$ to $A_\lambda$ has image precisely \begin{align*} \operatorname{SYT}(\lambda)\times \operatorname{SYT}(\lambda). \end{align*} Therefore the restricted map \begin{align*} \operatorname{RSK}|_{A_\lambda}:A_\lambda\to \operatorname{SYT}(\lambda)\times \operatorname{SYT}(\lambda) \end{align*} is a bijection. [guided] We isolate the fibre whose cardinality we want to compute. Define \begin{align*} A_\lambda:=\{w\in S_n:\operatorname{sh}(w)=\lambda\}. \end{align*} Thus the left-hand side of the desired identity is $|A_\lambda|$. Next define \begin{align*} \operatorname{SYT}(\lambda):=\{T:T \text{ is a standard Young tableau of shape } \lambda\}. \end{align*} By the meaning of $f_\lambda$, we have \begin{align*} f_\lambda=|\operatorname{SYT}(\lambda)|. \end{align*} Now define \begin{align*} \mathcal{C}_n:=\{(P,Q):\text{there exists }\mu\vdash n\text{ such that }P,Q\in\operatorname{SYT}(\mu)\} \end{align*} for the set of ordered pairs of standard Young tableaux of the same shape and total size $n$. Apply the RSK bijection for permutations, [citetheorem:8435]. Its hypotheses are satisfied because we are working with permutations in $S_n$, and the theorem states that the map $\operatorname{RSK}:S_n\to\mathcal{C}_n$ given by $w\mapsto (P(w),Q(w))$ is a bijection. The only point to check is what the condition $\operatorname{sh}(w)=\lambda$ becomes after applying RSK. By definition, $\operatorname{sh}(w)$ is the common shape of the insertion tableau $P(w)$ and the recording tableau $Q(w)$. Hence \begin{align*} \operatorname{sh}(w)=\lambda \end{align*} holds exactly when both $P(w)$ and $Q(w)$ have shape $\lambda$. Therefore the RSK image of $A_\lambda$ is precisely \begin{align*} \operatorname{SYT}(\lambda)\times \operatorname{SYT}(\lambda). \end{align*} Because the original RSK map on $S_n$ is bijective, its restriction to a subset is injective, and every pair in the displayed product has a unique preimage in $S_n$. Since that pair has shape $\lambda$, its preimage lies in $A_\lambda$. Thus \begin{align*} \operatorname{RSK}|_{A_\lambda}:A_\lambda\to \operatorname{SYT}(\lambda)\times \operatorname{SYT}(\lambda) \end{align*} is a bijection. [/guided] [/step] [step:Count ordered pairs of standard tableaux of shape $\lambda$] Since $\operatorname{RSK}|_{A_\lambda}$ is a bijection, finite sets have equal cardinality under it: \begin{align*} |A_\lambda|=|\operatorname{SYT}(\lambda)\times \operatorname{SYT}(\lambda)|. \end{align*} The two coordinates of an ordered pair are chosen independently from the same finite set $\operatorname{SYT}(\lambda)$, so \begin{align*} |\operatorname{SYT}(\lambda)\times \operatorname{SYT}(\lambda)|=|\operatorname{SYT}(\lambda)|^2. \end{align*} Using $|\operatorname{SYT}(\lambda)|=f_\lambda$, we obtain \begin{align*} |\{w\in S_n:\operatorname{sh}(w)=\lambda\}|=|A_\lambda|=(f_\lambda)^2. \end{align*} This is the required identity. [/step]