If we can show this after base change to the algebraic closure of $k$, then the result follows over $k$ because the kernel of pullback is torsion by Lemma \ref{lemma-kernel-to-closure}. Hence we may and do assume $k$ is algebraically closed. \medskip\noindent Using Bertini we can choose a smooth curve $C \subset X$ passing through $x$ and $x'$. See proof of Lemma \ref{lemma-divide-difference-points}. Hence we may assume $X$ is a curve. \medskip\noindent Assume $X$ is a curve and $k$ is algebraically closed. Write $S^n(X) = \underline{\Hilbfunctor}^n_{X/k}$ with notation as in Picard Schemes of Curves, Sections \ref{pic-section-hilbert-scheme-points} and \ref{pic-section-divisors}. There is a canonical morphism $$ \pi : X^n \longrightarrow S^n(X) $$ which sends the $k$-rational point $(x_1, \ldots, x_n)$ to the $k$-rational point corresponding to the divisor $[x_1] + \ldots + [x_n]$ on $X$. There is a faithful action of the symmetric group $S_n$ on $X^n$. The morphism $\pi$ is $S_n$-invariant and the fibres of $\pi$ are $S_n$-orbits (set theoretically). Finally, $\pi$ is finite flat of degree $n!$, see Picard Schemes of Curves, Lemma \ref{pic-lemma-universal-object}. \medskip\noindent Let $\alpha_n$ be the zero cycle on $X^n$ given by the formula in the statement of the lemma. Let $\mathcal{L} = \mathcal{O}_X(x - x')$. Then $c_1(\mathcal{L}) \cap [X] = [x] - [x']$. Thus $$ \alpha_n = c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_n) \cap [X^n] $$ where $\mathcal{L}_i = \text{pr}_i^*\mathcal{L}$ and $\text{pr}_i : X^n \to X$ is the $i$th projection. By either Divisors, Lemma \ref{divisors-lemma-finite-locally-free-has-norm} or Divisors, Lemma \ref{divisors-lemma-norm-in-normal-case} there is a norm for $\pi$. Set $\mathcal{N} = \text{Norm}_\pi(\mathcal{L}_1)$,  see Divisors, Lemma \ref{divisors-lemma-norm-invertible}. We have $$ \pi^*\mathcal{N} = (\mathcal{L}_1 \otimes \ldots \otimes \mathcal{L}_n)^{\otimes (n - 1)!} $$ in $\Pic(X^n)$ by a calculation. Deails omitted; hint: this follows from the fact that $\text{Norm}_\pi : \pi_*\mathcal{O}_{X^n} \to \mathcal{O}_{S^n(X)}$ composed with the natural map $\pi_*\mathcal{O}_{S^n(X)} \to \mathcal{O}_{X^n}$ is equal to the product over all $\sigma \in S_n$ of the action of $\sigma$ on $\pi_*\mathcal{O}_{X^n}$. Consider $$ \beta_n = c_1(\mathcal{N})^n \cap [S^n(X)] $$ in $\CH_0(S^n(X))$. Observe that $c_1(\mathcal{L}_i) \cap c_1(\mathcal{L}_i) = 0$ because $\mathcal{L}_i$ is pulled back from a curve, see Chow Homology, Lemma \ref{chow-lemma-vanish-above-dimension}. Thus we see that \begin{align*} \pi^*\beta_n & = ((n - 1)!)^n (\sum\nolimits_{i = 1, \ldots, n} c_1(\mathcal{L}_i))^n \cap [X^n] \\ & = ((n - 1)!)^n n^n  c_1(\mathcal{L}_1) \cap \ldots \cap c_1(\mathcal{L}_n) \cap [X^n] \\ & = (n!)^n \alpha_n \end{align*} Thus it suffices to show that $\beta_n$ is torsion. \medskip\noindent There is a canonical morphism $$ f : S^n(X) \longrightarrow \underline{\Picardfunctor}^n_{X/k} $$ See Picard Schemes of Curves, Lemma \ref{pic-lemma-picard-pieces}. For $n \geq 2g - 1$ this morphism is a projective space bundle (details omitted; compare with the proof of Picard Schemes of Curves, Lemma \ref{pic-lemma-picard-pieces}). The invertible sheaf $\mathcal{N}$ is trivial on the fibres of $f$, see below. Thus by the projective space bundle formula (Chow Homology, Lemma \ref{chow-lemma-chow-ring-projective-bundle}) we see that $\mathcal{N} = f^*\mathcal{M}$ for some invertible module $\mathcal{M}$ on $\underline{\Picardfunctor}^n_{X/k}$. Of course, then we see that $$ c_1(\mathcal{N})^n = f^*(c_1(\mathcal{M})^n) $$ is zero because $n > g = \dim(\underline{\Picardfunctor}^n_{X/k})$ and we can use Chow Homology, Lemma \ref{chow-lemma-vanish-above-dimension} as before. \medskip\noindent We still have to show that $\mathcal{N}$ is trivial on a fibre $F$ of $f$. Since the fibres of $f$ are projective spaces and since $\Pic(\mathbf{P}^m_k) = \mathbf{Z}$ (Divisors, Lemma \ref{divisors-lemma-Pic-projective-space-UFD}), this can be shown by computing the degree of $\mathcal{N}$ on a line contained in the fibre. Instead we will prove it by proving that $\mathcal{N}$ is algebraically equivalent to zero. First we claim there is a connected finite type scheme $T$ over $k$, an invertible module $\mathcal{L}'$ on $T \times X$ and $k$-rational points $p, q \in T$ such that $\mathcal{M}_p \cong \mathcal{O}_X$ and $\mathcal{M}_q = \mathcal{L}$. Namely, since $\mathcal{L} = \mathcal{O}_X(x - x')$ we can take $T = X$, $p = x'$, $q = x$, and $\mathcal{L}' = \mathcal{O}_{X \times X}(\Delta) \otimes \text{pr}_2^*\mathcal{O}_X(-x')$. Then we let $\mathcal{L}'_i$ on $T \times X^n$ for $i = 1, \ldots, n$ be the pullback of $\mathcal{L}'$ by $\text{id}_T \times \text{pr}_i : T \times X^n \to T \times X$. Finally, we let $\mathcal{N}' = \text{Norm}_{\text{id}_T \times \pi}(\mathcal{L}'_1)$ on $T \times S^n(X)$. By construction we have $\mathcal{N}'_p = \mathcal{O}_{S^n(X)}$ and $\mathcal{N}'_q = \mathcal{N}$. We conclude that $$ \mathcal{N}'|_{T \times F} $$ is an invertible module on $T \times F \cong T \times \mathbf{P}^m_k$ whose fibre over $p$ is the trivial invertible module and whose fibre over $q$ is $\mathcal{N}|_F$. Since the euler characteristic of the trivial bundle is $1$ and since this euler characteristic is locally constant in families (Derived Categories of Schemes, Lemma \ref{perfect-lemma-chi-locally-constant-geometric}) we conclude $\chi(F, \mathcal{N}^{\otimes s}|_F) = 1$ for all $s \in \mathbf{Z}$. This can happen only if $\mathcal{N}|_F \cong \mathcal{O}_F$ (see Cohomology of Schemes, Lemma \ref{coherent-lemma-cohomology-projective-space-over-ring}) and the proof is complete. Some details omitted.