Let $\mathcal{F}$ be a crystal in quasi-coherent modules on $X/S$. Set $T_e = \Spec(D_e)$ so that $(X, T_e, \bar\gamma)$ is an object of $\text{Cris}(X/S)$ for $e \gg 0$. We have morphisms $$ (X, T_e, \bar\gamma) \to (X, T_{e + 1}, \bar\gamma) \to \ldots $$ which are closed immersions. We set $$ M = \lim_e \Gamma((X, T_e, \bar\gamma), \mathcal{F}) = \lim_e \Gamma(T_e, \mathcal{F}_{T_e}) = \lim_e M_e $$ Note that since $\mathcal{F}$ is locally quasi-coherent we have $\mathcal{F}_{T_e} = \widetilde{M_e}$. Since $\mathcal{F}$ is a crystal we have $M_e = M_{e + 1}/p^eM_{e + 1}$. Hence we see that $M_e = M/p^eM$ and that $M$ is $p$-adically complete, see Algebra, Lemma \ref{algebra-lemma-limit-complete}. \medskip\noindent By Lemma \ref{lemma-automatic-connection} we know that $\mathcal{F}$ comes endowed with a canonical integrable connection $\nabla : \mathcal{F} \to \mathcal{F} \otimes \Omega_{X/S}$. If we evaluate this connection on the objects $T_e$ constructed above we obtain a canonical integrable connection $$ \nabla : M \longrightarrow M \otimes^\wedge_D \Omega_D $$ To see that this is topologically nilpotent we work out what this means. \medskip\noindent Now we can do the same procedure for the rings $D(n)$. This produces a $p$-adically complete $D(n)$-module $M(n)$. Again using the crystal property of $\mathcal{F}$ we obtain isomorphisms $$ M \otimes^\wedge_{D, p_0} D(1) \rightarrow M(1) \leftarrow M \otimes^\wedge_{D, p_1} D(1) $$ compare with the proof of Lemma \ref{lemma-automatic-connection}. Denote $c$ the composition from left to right. Pick $m \in M$. Write $\xi_i = x_i \otimes 1 - 1 \otimes x_i$. Using (\ref{equation-structure-Dn}) we can write uniquely $$ c(m \otimes 1) = \sum\nolimits_K \theta_K(m) \otimes \prod \xi_i^{[k_i]} $$ for some $\theta_K(m) \in M$ where the sum is over multi-indices $K = (k_i)$ with $k_i \geq 0$ and $\sum k_i < \infty$. Set $\theta_i = \theta_K$ where $K$ has a $1$ in the $i$th spot and zeros elsewhere. We have $$ \nabla(m) = \sum \theta_i(m) \text{d}x_i. $$ as can be seen by comparing with the definition of $\nabla$. Namely, the defining equation is $p_1^*m = \nabla(m) - c(p_0^*m)$ in Lemma \ref{lemma-automatic-connection} but the sign works out because in the Stacks project we consistently use $\text{d}f = p_1(f) - p_0(f)$ modulo the ideal of the diagonal squared, and hence $\xi_i = x_i \otimes 1 - 1 \otimes x_i$ maps to $-\text{d}x_i$ modulo the ideal of the diagonal squared. \medskip\noindent Denote $q_i : D \to D(2)$ and $q_{ij} : D(1) \to D(2)$ the coprojections corresponding to the indices $i, j$. As in the last paragraph of the proof of Lemma \ref{lemma-automatic-connection} we see that $$ q_{02}^*c = q_{12}^*c \circ q_{01}^*c. $$ This means that $$ \sum\nolimits_{K''} \theta_{K''}(m) \otimes \prod {\zeta''_i}^{[k''_i]} = \sum\nolimits_{K', K} \theta_{K'}(\theta_K(m)) \otimes \prod {\zeta'_i}^{[k'_i]} \prod \zeta_i^{[k_i]} $$ in $M \otimes^\wedge_{D, q_2} D(2)$ where \begin{align*} \zeta_i & = x_i \otimes 1 \otimes 1 - 1 \otimes x_i \otimes 1,\\ \zeta'_i & = 1 \otimes x_i \otimes 1 - 1 \otimes 1 \otimes x_i,\\ \zeta''_i & = x_i \otimes 1 \otimes 1 - 1 \otimes 1 \otimes x_i. \end{align*} In particular $\zeta''_i = \zeta_i + \zeta'_i$ and we have that $D(2)$ is the $p$-adic completion of the divided power polynomial ring in $\zeta_i, \zeta'_i$ over $q_2(D)$, see Lemma \ref{lemma-structure-Dn}. Comparing coefficients in the expression above it follows immediately that $\theta_i \circ \theta_j = \theta_j \circ \theta_i$ (this provides an alternative proof of the integrability of $\nabla$) and that $$ \theta_K(m) = (\prod \theta_i^{k_i})(m). $$ In particular, as the sum expressing $c(m \otimes 1)$ above has to converge $p$-adically we conclude that for each $i$ and each $m \in M$ only a finite number of $\theta_i^k(m)$ are allowed to be nonzero modulo $p$.