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. \medskip\noindent 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 \ref{algebra-theorem-universally-exact-criteria}. 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 \ref{algebra-lemma-Artin-Rees}) there exists a constant $c$ such that $\Im(R^{\oplus k} \to R^{\oplus m}) \cap (I^N)^{\oplus m} \subset \Im((I^{N - c})^{\oplus k} \to R^{\oplus m})$. 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.