I'm following the steps suggested in Rotman's text (Problem 2.30).
Let $R$ be a commutative ring and let $A_1, \ldots, A_n, M$ be $R$-modules with $n \ge 2$. Let $F$ be a free $R$-module with basis $\Pi A_i$ and define $N \subset F$ to be the submodule generated by $$ (a_1, \ldots, r a_i, \ldots, a_n) - r(a_1, \ldots, a_n) $$ and $$(\ldots, a_i + a'_i) - (\ldots, a_i, \ldots) - (\ldots, a'_i, \ldots).$$ Define $\bigotimes A_i := F/N$ and $h: \Pi A_i \to F/N $ by $(a_i) \mapsto (a_i) + N$.
Lemma
$h$ is $R$-multilinear and $h$ and $F/N$ solve the universal mapping problem for $R$-multilinear functions.
Proof. Given the way $N$ is defined, it is clear that $h$ is multilinear. Now consider the below commutative diagram
$i: \Pi A_i \to F$ is the inclusion, $\pi$ is the natural map, $f$ is any multilinear function.
$\phi$ exists by Proposition 2.34 (extending by linearity). By the definition of N and because $f$ is multilinear, $N \subset \text{ker} \phi$ so that it induces an $R$-map $\hat f : F/N \to M$ by $a \otimes b \mapsto \phi(a,b)$.
So we have $\hat f h = f$ as desired. $\square$
The categorical solution
Elsewhere in the text when talking about generalized associativity of the tensor product, he references a text that shows that any monoidal category has general associativity. I don't want to go that route here since it's not the approach Rotman wants.
Rotman's approach to generalized associativity
Problem 2.30(ii) is the above Lemma, so this suggests Rotman possibly wants to use this to prove general associativity which is part (iii) of the same problem. He gives this hint:
Hint. Prove that any association of $\mathcal A := A_1 \otimes \ldots \otimes A_n$ is also a solution to the universal mapping problem.
Questions
But what does an arbitrary association of $\mathcal A$ look like?
If we ignore nesting, we could use the Lemma and consider an association of the kind $\bigotimes_{p=1}^k \bigotimes_{i \in I_p} A_i $, where $I_p$ is a partition of $1 \ldots n$. I think this proof would be very similar to the one here.
But neither my type of partioned nesting nor the left-bracketed version in the linked post seems to prove "generalized" associativity (i.e. it doesn't address all cases).
