Unique factorization of free products of groups satisfying descending chain condition I am self-studying group theory, and proving Exercise 11.61 of Rotman's An Introduction to the Theory of Groups, on free products:

Let $A_1, \ldots, A_n, B_1, \ldots, B_m$ be indecomposable groups having both chain conditions. If $A_1 * \cdots * A_n \cong B_1 * \cdots * B_n$, then $n = m$ and there is a permutation $\sigma$ of $\{1,2,\ldots,n\}$ such that $B_{\sigma(i)} \cong A_i$ for all $i$.

(By "both chain conditions", it refers to the ascending and descending chain conditions on normal subgroups.)
Edit: the below answer does not rely on indecomposability or ascending chain condition, just the descending chain condition.
 A: There is this argument that does not even rely on $A_1, \ldots, A_n$ being indecomposable or satisfying ACC, just satisfying DCC:
By this question, we can construct an infinite descending chain of normal subgroups of any free product with at least two nontrivial factors: if $N(g)$ denotes the normal subgroup generated by an element $g$, let $x$ and $y$ be nontrivial elements from two distinct factors of the free product, and consider:
$$N(xy) > N((xy)^2) > N((xy)^4) > \ldots > N((xy)^{2^n}) > \ldots$$
This proves that any free product with at least two nontrivial factors does not satisfy the DCC. As well, $F_1 \cong \mathbb{Z}$ does not satisfy the DCC, so no nontrivial free group satisfies the DCC. Therefore by Kurosh's theorem (which is in the next section after this exercise, but I'll take it!), the maximal subgroups of $A_1 * \cdots * A_n$ satisfying the DCC are precisely $A_1, \ldots, A_n$ and their conjugates. Partition the collection of these maximal such subgroups into equivalence classes for if they are conjugate, and pick one group from every equivalence class. We recover $A_1, \ldots, A_n$ up to isomorphism in some order.
