= Solution
For $k=1$, the product is empty and equals the identity, the only $1$-cycle in $S_1$. Suppose the <cycle-sum identity for Young–Jucys–Murphy elements> holds at $k-1$:
$$
X_2\cdots X_{k-1}=\sum_{\sigma\text{ a }(k-1)\text{-cycle}}\sigma.
$$
Multiplication by $X_k=\sum_{a<k}(a\ k)$ gives terms $\sigma(a\ k)$. With permutation products acting right to left, this inserts $k$ immediately after $a$ in the cycle of $\sigma$: the arrows become $a\mapsto k\mapsto\sigma(a)$, with the other arrows unchanged.
Conversely, in any $k$-cycle there is a unique predecessor $a$ of $k$. Delete $k$ and join that predecessor to the successor of $k$; this uniquely recovers $\sigma$ and the factor $(a\ k)$. Hence every $k$-cycle occurs exactly once, and no other permutation occurs. Induction proves
$$
\boxed{\xi_k=X_2\cdots X_k=\sum_{\sigma\text{ a }k\text{-cycle in }S_k}\sigma}.
$$
Thus $\xi_k$ is a <conjugacy-class sum> in $\mathbb C S_k$, although it need not be central in $\mathbb C S_n$ when $k<n$.
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