**Proof.**
Assume $g \geq 2$. First we may choose a finite separable extension of degree at most $2g - 2$ such that $X$ acquires a rational point, see Lemma 53.13.9. Thus we may assume $X$ has a $k$-rational point $x \in X(k)$ but now we have to prove the lemma with $[k' : k] \leq (n^{2g})!$. Let $k \subset k^{sep} \subset \overline{k}$ be a separable algebraic closure inside an algebraic closure. By Lemma 53.17.1 we have

\[ \mathop{\mathrm{Pic}}\nolimits (X_{\overline{k}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g} \]

By Picard Schemes of Curves, Lemma 44.7.2 we conclude that

\[ \mathop{\mathrm{Pic}}\nolimits (X_{k^{sep}})[n] \cong (\mathbf{Z}/n\mathbf{Z})^{\oplus 2g} \]

By Picard Schemes of Curves, Lemma 44.7.2 there is a continuous action

\[ \text{Gal}(k^{sep}/k) \longrightarrow \text{Aut}(\mathop{\mathrm{Pic}}\nolimits (X_{k^{sep}})[n] \]

and the lemma is true for the fixed field $k'$ of the kernel of this map. The kernel is open because the action is continuous which implies that $k'/k$ is finite. By Galois theory $\text{Gal}(k'/k)$ is the image of the displayed arrow. Since the permutation group of a set of cardinality $n^{2g}$ has cardinality $(n^{2g})!$ we conclude by Galois theory that $[k' : k] \leq (n^{2g})!$. (Of course this proves the lemma with the bound $|\text{GL}_{2g}(\mathbf{Z}/n\mathbf{Z})|$, but all we want here is that there is some bound.)

If the genus is $1$, then there is no upper bound on the degree of a finite separable field extension over which $X$ acquires a rational point (details omitted). Still, there is such an extension for example by Varieties, Lemma 33.25.6. The rest of the proof is the same as in the case of $g \geq 2$.
$\square$

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