Faithfully flat descent for modules from the simplicial point of view Let $R \rightarrow R'$ be a faithfully flat ring map, let $M$ be an $R$-module, and let $M_n$ be the base change of $M$ to the tensor product of $n + 1$ copies of $R'$ over $R$. One way to formulate the statement of faithfully flat descent is as the claim that the complex 
$$ 0 \rightarrow M \rightarrow M_0 \rightarrow M_1 \rightarrow \dotsc$$
is exact, where the maps are the Cech style alternating sums of pullbacks. This is a fairly common way that is discussed further here https://stacks.math.columbia.edu/tag/023M.
Another, probably less common, way to formulate faithfully flat descent is to say that 
$$ M \xrightarrow{\sim} Rlim_{\Delta} M_n$$
where $\Delta$ is the usual simplex category that is used to define simplicial objects. 
My question is: why are these two formulations equivalent? How does one see that one implies the other?
I suspect that one probably uses the Dold-Kan equivalence and then some "standard" but possibly difficult to locate in the literature simplicial calculus. I am guessing that the core of my question is not really about descent for modules, but rather about simplicial or bisimplicial objects and Rlim. I would very much appreciate if someone could explain the equivalence between the two points of view in detail (or give a precise reference to the literature where this is treated).
 A: Essentially, your question is: for any co-simplicial abelian group $M_{\bullet}$, why is ${\rm R lim}_{\Delta} M_{\bullet}$ given by the complex  $$C:= M_0 \to M_1 \to M_2 \to \dots ?$$ (If $A$ is an abelian group $A \to C$ is a quasi-isomorphism if and only if $M \to M_0 \to M_1 \to M_2 \to \dots $ is acyclic).
There is a direct lowbrow proof, given by resolving the constant cosimplicial abelian group $\mathbb Z *$  given by $\mathbb Z *_n = \mathbb Z$ and all maps the identity.   Since we have a natural isomorphism $${\rm Hom}_{\Delta}(\mathbb Z *, M_{\bullet}) \cong {\rm lim}_{\Delta} M_{\bullet}$$ there is a quasi-isomorphism  ${\rm RHom}_{\Delta}(\mathbb Z *, M_{\bullet}) \simeq {\rm R lim}_{\Delta} M_{\bullet}$. To compute it, we resolve  $\mathbb Z *$ by free (or if you prefer, representable) co-simplicial abelian groups. 
The resolution looks like $$\mathbb Z \Delta(0,-) \leftarrow \mathbb Z \Delta (1, -) \leftarrow \mathbb Z \Delta(2, -) \leftarrow \mathbb Z \Delta(3,-) \leftarrow \dots $$
Here, the number $n$ represents the totally ordered set $\{0, \dots, n\}$  and the cosimplicial abelian group  $\mathbb Z \Delta(n, -)$ is given by $\mathbb Z \Delta(n, -)_m = \mathbb Z \Delta(n, m)$.  The differential $\mathbb Z \Delta(n, -) \to \mathbb Z \Delta(n-1, -)$  is given by the alternating sum of precomposition by the $n+1$ face maps in $\Delta(n-1, n)$.  
The complex resolves  $\mathbb Z *$,  because if we look at level $n$,  we see that we get exactly the simplicial homology complex for the $n$-simplex!  $$\mathbb Z \Delta(0,n) \leftarrow \mathbb Z \Delta(1,n) \leftarrow \dots$$ Since we know the homology of the simplex,  we know that the complex is a resolution.  
Now by the Yoneda lemma,  we have that ${\rm Hom}_{\Delta}( \mathbb Z \Delta(n,-), M) = M_n$.  In particular,  this resolution is projective.  Using it to compute ${\rm RHom}(\mathbb Z *, M)$ we get the complex  $${\rm Hom}_{\Delta}( \mathbb Z \Delta(0,-), M) \to {\rm Hom}_{\Delta}( \mathbb Z \Delta(1,-), M) \to {\rm Hom}_{\Delta}( \mathbb Z \Delta(2,-), M) \to \dots, $$  and applying Yoneda, this is exactly the complex $C$ we wanted.
