The hook-product ratios sum to the size

The numbers $h_{i,1}={\lambda }_i+r-i$ are strictly decreasing in $i$ and positive, because ${\lambda }_i\ge {\lambda }_{i+1}$ gives $h_{i,1}-h_{i+1,1}={\lambda }_i-{\lambda }_{i+1}+1\ge 1$ and $h_{r,1}={\lambda }_r\ge 1$; for $x=(a,b)\in \mathrm{Rem}(\lambda )$ we have $b={\lambda }_a$ and ${\lambda }_b^{\mathrm{\prime }}=a$. By [F1] the value $R(x)$ is a finite product of ratios $h_{\lambda }(y)\mathrm{/}(h_{\lambda }(y)-1)$ over $R_x$, which may be empty (for example when $\lambda =(1)$), and the sum over $\mathrm{Rem}(\mathrm{\varnothing })=\mathrm{\varnothing }$ is $0=n$.

The multiset identity for row $a$: the $h_{a,1}={\lambda }_a+(r-a)$ numbers of $L:=\{h_{\lambda }(a,j):1\le j\le {\lambda }_a\}\cup \{h_{a,1}-h_{k,1}:a<k\le r\}$ are pairwise distinct and all lie in $\{1,2,\dots ,h_{a,1}\}$, so as multisets $L=\{1,\dots ,h_{a,1}\}$. Indeed $h_{\lambda }(a,j)={\lambda }_a-j+{\lambda }_j^{\mathrm{\prime }}-a+1$ decreases strictly with $j$; the differences equal ${\lambda }_a-{\lambda }_k+k-a$ and increase strictly with $k$, lying between $1$ and $h_{a,1}-1$; and a repetition $h_{\lambda }(a,j)=h_{a,1}-h_{k,1}$ would force ${\lambda }_j^{\mathrm{\prime }}+{\lambda }_k=j+k-1$, which is impossible: if ${\lambda }_k\le j-1$ then the hook of $(a,j)$ does not reach row $k$, so ${\lambda }_j^{\mathrm{\prime }}<k$ and ${\lambda }_j^{\mathrm{\prime }}+{\lambda }_k<j+k-1$, while if ${\lambda }_k\ge j$ then $(k,j)$ lies in the column of $(a,j)$, so ${\lambda }_j^{\mathrm{\prime }}\ge k$ and ${\lambda }_j^{\mathrm{\prime }}+{\lambda }_k>j+k-1$.

Finite identity: for pairwise distinct $z_1,\dots ,z_r$ in a field $K$ with $r\ge 1$, ${\sum }_{i=1}^r{z}_i{\prod }_{j\mathrm{\ne }i}(1+\frac{1}{z_j-z_i})={\sum }_{i=1}^r{z}_i-\left(\genfrac{}{}{0px}{}{r}{2}\right)\mathrm{.}$ For $r=1$ the identity is $z_1=z_1$. Hence assume $r\ge 2$ for its coefficient calculation. Set $Q(t):={\prod }_{j=1}^r(t-z_j)\in K[t]$ and for $g\in K[t]$ whose coefficients above $t^{r-1}$ vanish let $\mathrm{\Lambda }(g):={\sum }_{i}g(z_i)\mathrm{/}{\prod }_{j\mathrm{\ne }i}(z_i-z_j)$. Then $\mathrm{\Lambda }(g)=[t^{r-1}]g$: the polynomial $g(t)-{\sum }_{i}g(z_i){\prod }_{j\mathrm{\ne }i}(t-z_j)\mathrm{/}(z_i-z_j)$ has no nonzero coefficient above $t^{r-1}$ and vanishes at $z_1,\dots ,z_r$, hence is zero by [F5], and comparing coefficients of $t^{r-1}$ gives the claim.

The column-$b$ factors of $R(x)$: for $1\le i<a$ the box $(i,b)$ lies in $[\mu ]$, and $h_{\lambda }(i,b)={\lambda }_i-b+{\lambda }_b^{\mathrm{\prime }}-i+1=h_{i,1}-h_{a,1}+1$, because ${\lambda }_b^{\mathrm{\prime }}=a$ and $h_{a,1}=b+r-a$; hence ${\prod }_{i<a}\frac{h_{\lambda }(i,b)}{h_{\lambda }(i,b)-1}={\prod }_{i<a}(1+\frac{1}{h_{i,1}-h_{a,1}})\mathrm{.}$

The row-$a$ factors of $R(x)$: if $b\ge 2$, applying step 1.2 to $\lambda$ and to $\mu =\lambda -x$ (which then has $r$ parts, row $a$ of length ${\lambda }_a-1$, and first-column hooks $h_{i,1}$ for $i\mathrm{\ne }a$, $h_{a,1}-1$ for $i=a$) and multiplying the two identities gives $({\prod }_{j<b}{h}_{\lambda }(a,j))\cdot {\prod }_{k>a}(h_{a,1}-h_{k,1})=h_{a,1}!,({\prod }_{j<b}(h_{\lambda }(a,j)-1))\cdot {\prod }_{k>a}(h_{a,1}-h_{k,1}-1)=(h_{a,1}-1)!,$ where $h_{\lambda }(a,b)=1$ by removability; dividing them yields ${\prod }_{j<b}\frac{h_{\lambda }(a,j)}{h_{\lambda }(a,j)-1}=h_{a,1}{\prod }_{k>a}\frac{h_{a,1}-h_{k,1}-1}{h_{a,1}-h_{k,1}}=h_{a,1}{\prod }_{k>a}(1+\frac{1}{h_{k,1}-h_{a,1}}),$ since $1+(h_{k,1}-h_{a,1}{)}^{-1}=(h_{a,1}-h_{k,1}-1)\mathrm{/}(h_{a,1}-h_{k,1})$ for $k>a$. If $b=1$ then ${\lambda }_a=1$ forces $a=r$, the product over $j<b$ is empty and $h_{a,1}=1$, so the same displayed formula holds trivially.

Set $g(t):=tQ(t-1)-(t-r)Q(t)=t(Q(t-1)-Q(t))+rQ(t)$. Both $Q(t)$ and $Q(t-1)$ are monic of degree $r$, so $[t^{r+1}]g=0$. Writing $e_1:={\sum }_j{z}_j$, their $t^{r-1}$ coefficients differ by $-r$, while $[t^r](rQ(t))=r$; hence $[t^r]g=-r+r=0$. Thus all coefficients of $g$ above $t^{r-1}$ vanish, including when $r=0$ in $K$ or $g=0$. Moreover, expanding $Q(t-1)={\prod }_j(t-(z_j+1))$ gives $[t^{r-2}](Q(t-1)-Q(t))=(r-1)e_1+\left(\genfrac{}{}{0px}{}{r}{2}\right)$ and $[t^{r-1}](rQ(t))=-r{e}_1$, so $[t^{r-1}]g=\left(\genfrac{}{}{0px}{}{r}{2}\right)-e_1$. Here integers are mapped into $K$, so no division by $2$ in $K$ is used.

Combining steps 2.1 and 2.2 with [F1], for every $x=(a,b)\in \mathrm{Rem}(\lambda )$, $R(x)=h_{a,1}{\prod }_{i\mathrm{\ne }a}(1+\frac{1}{h_{i,1}-h_{a,1}})\mathrm{.}$

Since $Q(z_i)=0$ and $Q(z_i-1)={\prod }_j(z_i-1-z_j)=-(-1{)}^{r-1}{\prod }_{j\mathrm{\ne }i}(z_j-z_i+1)$, for each $i$ $\frac{g(z_i)}{{\prod }_{j\mathrm{\ne }i}(z_i-z_j)}=\frac{z_{i}Q(z_i-1)}{{\prod }_{j\mathrm{\ne }i}(z_i-z_j)}=-z_i{\prod }_{j\mathrm{\ne }i}(1+\frac{1}{z_j-z_i})\mathrm{.}$ Summing over $i$ and using $\mathrm{\Lambda }(g)=[t^{r-1}]g$ from step 1.3 together with step 2.3 gives the finite identity.

Rows without removable nodes contribute zero and the sum may be extended over all rows: by [F3] the removable nodes correspond to the indices $a$ with ${\lambda }_a>{\lambda }_{a+1}$, and if $a<r$ satisfies ${\lambda }_a={\lambda }_{a+1}$, then $h_{a,1}-h_{a+1,1}=1$ and the factor of index $a+1$ in the product of step 3.1 vanishes, so the corresponding term is $0$. Therefore ${\sum }_{x\in \mathrm{Rem}(\lambda )}R(x)={\sum }_{a=1}^r{h}_{a,1}{\prod }_{i\mathrm{\ne }a}(1+\frac{1}{h_{i,1}-h_{a,1}})\mathrm{.}$

Applying the finite identity of steps 1.3 and 3.2 (the $r=1$ case being immediate in step 1.3) over $K=\mathbb{Q}$ to the pairwise distinct numbers $z_i:=h_{i,1}$ (step 1.1) gives ${\sum }_{a=1}^r{h}_{a,1}{\prod }_{i\mathrm{\ne }a}(1+\frac{1}{h_{i,1}-h_{a,1}})={\sum }_{i=1}^r{h}_{i,1}-\left(\genfrac{}{}{0px}{}{r}{2}\right)\mathrm{.}$

The first-column hooks sum to $n+\left(\genfrac{}{}{0px}{}{r}{2}\right)$: ${\sum }_i{h}_{i,1}={\sum }_i{\lambda }_i+{\sum }_i(r-i)=n+r(r-1)\mathrm{/}2=n+\left(\genfrac{}{}{0px}{}{r}{2}\right)$. Substituting this into step 4.2 and using step 4.1 yields ${\sum }_{x\in \mathrm{Rem}(\lambda )}R(x)=n$, and the case $\lambda =\mathrm{\varnothing }$ is the empty sum $0$; this proves the lemma.