Proof.
If $n = \ell n'$ for some prime number $\ell $, then we get a short exact sequence $0 \to M[\ell ] \to M \to M' \to 0$ of finite $\Lambda $-modules and $M'$ is annihilated by $n'$. This produces a corresponding short exact sequence of constant sheaves, which in turn gives rise to an exact sequence of cohomology modules
\[ H^ q_{\acute{e}tale}(X, \underline{M[n]}) \to H^ q_{\acute{e}tale}(X, \underline{M}) \to H^ q_{\acute{e}tale}(X, \underline{M'}) \]
Thus, if we can show the result in case $M$ is annihilated by a prime number, then by induction on $n$ we win.
Let $\ell $ be a prime number such that $\ell $ annihilates $M$. Then we can replace $\Lambda $ by the $\mathbf{F}_\ell $-algebra $\Lambda /\ell \Lambda $. Namely, the cohomology of $\mathcal{F}$ as a sheaf of $\Lambda $-modules is the same as the cohomology of $\mathcal{F}$ as a sheaf of $\Lambda /\ell \Lambda $-modules, for example by Cohomology on Sites, Lemma 21.12.4.
Assume $\ell $ be a prime number such that $\ell $ annihilates $M$ and $\Lambda $. Let us reduce to the case where $M$ is a finite free $\Lambda $-module. Namely, choose a short exact sequence
\[ 0 \to N \to \Lambda ^{\oplus m} \to M \to 0 \]
This determines an exact sequence
\[ H^ q_{\acute{e}tale}(X, \underline{\Lambda ^{\oplus m}}) \to H^ q_{\acute{e}tale}(X, \underline{M}) \to H^{q + 1}_{\acute{e}tale}(X, \underline{N}) \]
By descending induction on $q$ we get the result for $M$ if we know the result for $\Lambda ^{\oplus m}$. Here we use that we know that our cohomology groups vanish in degrees $> 2$ by Theorem 59.83.10.
Let $\ell $ be a prime number and assume that $\ell $ annihilates $\Lambda $. It remains to show that the cohomology groups $H^ q_{\acute{e}tale}(X, \underline{\Lambda })$ are finite $\Lambda $-modules. We will use a trick to show this; the “correct” argument uses a coefficient theorem which we will show later. Choose a basis $\Lambda = \bigoplus _{i \in I} \mathbf{F}_\ell e_ i$ such that $e_0 = 1$ for some $0 \in I$. The choice of this basis determines an isomorphism
\[ \underline{\Lambda } = \bigoplus \underline{\mathbf{F}_\ell } e_ i \]
of sheaves on $X_{\acute{e}tale}$. Thus we see that
\[ H^ q_{\acute{e}tale}(X, \underline{\Lambda }) = H^ q_{\acute{e}tale}(X, \bigoplus \underline{\mathbf{F}_\ell } e_ i) = \bigoplus H^ q_{\acute{e}tale}(X, \underline{\mathbf{F}_\ell })e_ i \]
since taking cohomology over $X$ commutes with direct sums by Theorem 59.51.3 (or Lemma 59.51.4 or Lemma 59.52.2). Since we already know that $H^ q_{\acute{e}tale}(X, \underline{\mathbf{F}_\ell })$ is a finite dimensional $\mathbf{F}_\ell $-vector space (by Theorem 59.83.10), we see that $H^ q_{\acute{e}tale}(X, \underline{\Lambda })$ is free over $\Lambda $ of the same rank. Namely, given a basis $\xi _1, \ldots , \xi _ m$ of $H^ q_{\acute{e}tale}(X, \underline{\mathbf{F}_\ell })$ we see that $\xi _1 e_0, \ldots , \xi _ m e_0$ form a $\Lambda $-basis for $H^ q_{\acute{e}tale}(X, \underline{\Lambda })$.
$\square$
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