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I came up with a more or less elementary proof of the identity from the top comment to my question. It involves nothing more advanced than some basic linear algebra.

Namely, let us denote $X_k=\sum\limits_{i=1}^{k-1} (i,k) \in\mathbb{C}[S_n]$, then we are to prove $$\prod\limits_{i=-k+1}^{k-1}(X_k-i)=0$$ for all $1\le k\le n$. For convenience sake we will also use $X_k$ to denote the linear operator on $\mathbb{C}[S_n]$ of left multiplication by $X_k \in \mathbb{C}[S_n]$.

First of all, let us show that $X_k$ is a diagonalizable operator. There are many ways to prove this fact, for example it is easy to see that $X_k$'s matrix is symmetric in the standard basis consisting of all the elements of $S_n$ (since the matrix corresponding to any $(i,k)$ is obviously such).

With the diagonalizability taken into account it is now sufficient to prove that $X_k$'s spectrum is a subset of $\{-k+1,-k+2,\ldots,k-1\}$. Starting with $X_1=0$ we conduct by induction on $k$.

Suppose that $\lambda\not\in\{-k,\ldots,k\}$ is a an eigenvalue of $X_{k+1}$. $X_{k+1}$ commutes with all of $\mathbb{C}[S_k]$ including $X_k$, which implies that $X_k$ and $X_{k+1}$ are simultaneously diagonalizable. Thus exists such a nonzero $v\in\mathbb{C}[S_n]$ that $X_{k+1}v=\lambda v$ and $X_kv=\mu v$. Our choice of $\lambda$ together with the inductive hypothesis provides $(\lambda-\mu)\not\in\{-1,0,1\}$ which lets us consider the element $$u=\left(s_k-\frac1{\lambda-\mu}\right)v$$ where $s_k=(k,k+1)$. $u\neq0$, otherwise we would have $s_kv=\frac1{\lambda-\mu}v\implies \lambda-\mu=\pm1$ since $s_k^2v=v$. Finally $$X_ku=X_ks_kv-\frac1{\lambda-\mu}X_kv=(s_kX_{k+1}-1)v-\frac\mu{\lambda-\mu}v=\lambda s_kv-v-\frac\mu{\lambda-\mu} v=\lambda u$$ where we employ the easily obtainable $s_kX_{k+1}=X_ks_k+1$. However $\lambda$ being an eigenvalue of $X_k$ contradicts the inductive hypothesis due to our choice of $\lambda$.

2 deleted 4 characters in body

I came up with a more or less elementary proof of the identity from the top comment to my question. It involves nothing more advanced than some basic linear algebra.

Namely, let us denote $X_k=\sum\limits_{i=1}^{k-1} (i,k) \in\mathbb{C}[S_n]$, then we are to prove $$\prod\limits_{i=-k+1}^{k-1}(X_k-i)=0$$ for all $1\le k\le n$. For convenience sake we will also use $X_k$ to denote the linear operator on $\mathbb{C}[S_n]$ of left multiplication by $X_k \in \mathbb{C}[S_n]$.

First of all, let us show that $X_k$ is a diagonalizable operator. There are many ways to prove this fact, for example it is easy to see that $X_k$'s matrix is symmetric in the standard basis consisting of all the elements of $S_n$ (since the matrix corresponding to any $(i,k)$ is obviously such).

With the diagonalizability taken into account it is now sufficient to prove that $X_k$'s spectrum is a subset of $\{-k+1,-k+2,\ldots,k-1\}$. Starting with $X_1=0$ we conduct by induction on $k$.

Suppose that $\lambda\not\in\{-k+1,\ldots,k-1\}$ \lambda\not\in\{-k,\ldots,k\}$is a an eigenvalue of$X_{k+1}$.$X_{k+1}$commutes with all of$\mathbb{C}[S_k]$including$X_k$, which implies that$X_k$and$X_{k+1}$are simultaneously diagonalizable. Thus exists such a nonzero$v\in\mathbb{C}[S_n]$that$X_{k+1}v=\lambda v$and$X_kv=\mu v$. Our choice of$\lambda$together with the inductive hypothesis provides$(\lambda-\mu)\not\in\{-1,0,1\}$which lets us consider the element $$u=\left(s_k-\frac1{\lambda-\mu}\right)v$$ where$s_k=(k,k+1)$.$u\neq0$, otherwise we would have$s_kv=\frac1{\lambda-\mu}v\implies \lambda-\mu=\pm1$since$s_k^2v=v$. Finally $$X_ku=X_ks_kv-\frac1{\lambda-\mu}X_kv=(s_kX_{k+1}-1)v-\frac\mu{\lambda-\mu}v=\lambda s_kv-v-\frac\mu{\lambda-\mu} v=\lambda u$$ where we employ the easily obtainable$s_kX_{k+1}=X_ks_k+1$. However$\lambda$being an eigenvalue of$X_k$contradicts the inductive hypothesis due to our choice of$\lambda$. 1 I came up with a more or less elementary proof of the identity from the top comment to my question. It involves nothing more advanced than some basic linear algebra. Namely, let us denote$X_k=\sum\limits_{i=1}^{k-1} (i,k) \in\mathbb{C}[S_n]$, then we are to prove $$\prod\limits_{i=-k+1}^{k-1}(X_k-i)=0$$ for all$1\le k\le n$. For convenience sake we will also use$X_k$to denote the linear operator on$\mathbb{C}[S_n]$of left multiplication by$X_k \in \mathbb{C}[S_n]$. First of all, let us show that$X_k$is a diagonalizable operator. There are many ways to prove this fact, for example it is easy to see that$X_k$'s matrix is symmetric in the standard basis consisting of all the elements of$S_n$(since the matrix corresponding to any$(i,k)$is obviously such). With the diagonalizability taken into account it is now sufficient to prove that$X_k$'s spectrum is a subset of$\{-k+1,-k+2,\ldots,k-1\}$. Starting with$X_1=0$we conduct by induction on$k$. Suppose that$\lambda\not\in\{-k+1,\ldots,k-1\}$is a an eigenvalue of$X_{k+1}$.$X_{k+1}$commutes with all of$\mathbb{C}[S_k]$including$X_k$, which implies that$X_k$and$X_{k+1}$are simultaneously diagonalizable. Thus exists such a nonzero$v\in\mathbb{C}[S_n]$that$X_{k+1}v=\lambda v$and$X_kv=\mu v$. Our choice of$\lambda$together with the inductive hypothesis provides$(\lambda-\mu)\not\in\{-1,0,1\}$which lets us consider the element $$u=\left(s_k-\frac1{\lambda-\mu}\right)v$$ where$s_k=(k,k+1)$.$u\neq0$, otherwise we would have$s_kv=\frac1{\lambda-\mu}v\implies \lambda-\mu=\pm1$since$s_k^2v=v$. Finally $$X_ku=X_ks_kv-\frac1{\lambda-\mu}X_kv=(s_kX_{k+1}-1)v-\frac\mu{\lambda-\mu}v=\lambda s_kv-v-\frac\mu{\lambda-\mu} v=\lambda u$$ where we employ the easily obtainable$s_kX_{k+1}=X_ks_k+1$. However$\lambda$being an eigenvalue of$X_k$contradicts the inductive hypothesis due to our choice of$\lambda\$.