**Proof.**
By definition an element $x$ of the left hand side is $x = (x_ n)$ where $x_ n = (x_{n, \alpha }) \in \bigoplus \nolimits _{\alpha \in A} R/I^ n$ such that $x_{n, \alpha } = x_{n + 1, \alpha } \bmod I^ n$. As $R = R^\wedge $ we see that for any $\alpha $ there exists a $y_\alpha \in R$ such that $x_{n, \alpha } = y_\alpha \bmod I^ n$. Note that for each $n$ there are only finitely many $\alpha $ such that the elements $x_{n, \alpha }$ are nonzero. Conversely, given $(y_\alpha ) \in \prod _\alpha R$ such that for each $n$ there are only finitely many $\alpha $ such that $y_{\alpha } \bmod I^ n$ is nonzero, then this defines an element of the left hand side. Hence we can think of an element of the left hand side as infinite “convergent sums” $\sum _\alpha y_\alpha $ with $y_\alpha \in R$ such that for each $n$ there are only finitely many $y_\alpha $ which are nonzero modulo $I^ n$. The displayed map maps this element to the element to $(y_\alpha )$ in the product. In particular the map is injective.

Let $Q$ be a finite $R$-module. We have to show that the map

\[ Q \otimes _ R \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \longrightarrow Q \otimes _ R \left(\prod \nolimits _{\alpha \in A} R\right) \]

is injective, see Algebra, Theorem 10.82.3. Choose a presentation $R^{\oplus k} \to R^{\oplus m} \to Q \to 0$ and denote $q_1, \ldots , q_ m \in Q$ the corresponding generators for $Q$. By Artin-Rees (Algebra, Lemma 10.51.2) there exists a constant $c$ such that $\mathop{\mathrm{Im}}(R^{\oplus k} \to R^{\oplus m}) \cap (I^ N)^{\oplus m} \subset \mathop{\mathrm{Im}}((I^{N - c})^{\oplus k} \to R^{\oplus m})$. Let us contemplate the diagram

\[ \xymatrix{ \bigoplus _{l = 1}^ k \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & \bigoplus _{j = 1}^ m \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & Q \otimes _ R \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge \ar[r] \ar[d] & 0 \\ \bigoplus _{l = 1}^ k \left(\prod \nolimits _{\alpha \in A} R\right) \ar[r] & \bigoplus _{j = 1}^ m \left(\prod \nolimits _{\alpha \in A} R\right) \ar[r] & Q \otimes _ R \left(\prod \nolimits _{\alpha \in A} R\right) \ar[r] & 0 } \]

with exact rows. Pick an element $\sum _ j \sum _\alpha y_{j, \alpha }$ of $\bigoplus _{j = 1, \ldots , m} \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge $. If this element maps to zero in the module $Q \otimes _ R \left(\prod \nolimits _{\alpha \in A} R\right)$, then we see in particular that $\sum _ j q_ j \otimes y_{j, \alpha } = 0$ in $Q$ for each $\alpha $. Thus we can find an element $(z_{1, \alpha }, \ldots , z_{k, \alpha }) \in \bigoplus _{l = 1, \ldots , k} R$ which maps to $(y_{1, \alpha }, \ldots , y_{m, \alpha }) \in \bigoplus _{j = 1, \ldots , m} R$. Moreover, if $y_{j, \alpha } \in I^{N_\alpha }$ for $j = 1, \ldots , m$, then we may assume that $z_{l, \alpha } \in I^{N_\alpha - c}$ for $l = 1, \ldots , k$. Hence the sum $\sum _ l \sum _\alpha z_{l, \alpha }$ is “convergent” and defines an element of $\bigoplus _{l = 1, \ldots , k} \left(\bigoplus \nolimits _{\alpha \in A} R\right)^\wedge $ which maps to the element $\sum _ j \sum _\alpha y_{j, \alpha }$ we started out with. Thus the right vertical arrow is injective and we win.
$\square$

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