Aluffi III.6.8 suggests proving that, for $R$ a ring and $A$ any set, $H = \text{Hom}_R(R^{\oplus A}, R)$ satisfies the universal property for the product of the familiy of $\{R_a\}_{a \in A}$, where $R_a \cong R$ for all $a$.
This question is already asked and partially answered here and here, but since I tried proving that myself before looking on math.SE and got a slightly different proof (one that shows the universal property explicitly, perhaps analogously to the second linked question, but I cannot see this easily), I'm curious if my version is correct.
So, define projection $\pi_a : H \rightarrow R_a$ by $\pi_a(\varphi) = \varphi(I_a)$, where $I_a \in R^{\oplus A}$ is the indicator function ($I_a(x) = 1$ if $x = a$ and $0$ otherwise)1. It's easy to verify that $\pi_a$ is in fact an $R$-module homomorphism.
Then, take arbitrary $R$-module $N$ and a collection of homomorphisms $\varphi_a : N \rightarrow R_a$. We need to define $\sigma : N \rightarrow H$ and show that it's unique.
Commutativity of the corresponding diagram forces $\forall a : A. \forall n : N. \pi_a (\sigma_n) = \varphi_a(n)$, or, expanding the definition for projections, $\sigma_n(I_a) = \varphi_a(n)$. Now, any morphism in $R^{\oplus A}$ can be defined as a finite sum of $I_a$, and the homomorphism condition forces $\sigma_n$ to be defined accordingly, so it exists and is unique, as desired.
Does this look reasonable?
1 Why this specific argument to $\varphi$? Frankly I don't know, it just feels like a good candidate based on the structure of the proofs I've encountered and built so far — it's a simplest nontrivial function useful as a building block of sorts.