Let $H$ be a $\mathbb R$-Hilbert space and $A_n\in\mathfrak L(H)$ be compact for $n\in\mathbb N$.
How can we determine the singular values of the product $B_n:=A_n\cdots A_1$?
We know that $$A_n^\ast A_n=\sum_{i\in I_n}\left|\lambda_i^{(n)}\right|^2e_i^{(n)}\otimes e_i^{(n)}\tag1,$$ where $I_n:=\mathbb N\cap[0,\operatorname{rank}(A_n^\ast A_n)]$, $\left(\lambda_i^{(n)}\right)_{i\in I_n}\subseteq(0,\infty)$ is nonincreasing and $\left(e_i^{(n)}\right)_{i\in I_n}$ is an orthonormal basis of$^1$ $\overline{\mathcal R(A_n^\ast A)}$. From $(1)$ it's easy to see that $$|A_n|:=\sqrt{A_n^\ast A_n}=\sum_{i\in I_n}\lambda_i^{(n)}e_i^{(n)}\otimes e_i^{(n)}\tag2.$$
By definition, the singular values of $A_n$ are precisely given by $\left(\lambda_i^{(n)}\right)_{i\in I_n}$.
$^1$ Note that $\mathcal N(|A_n|)=\mathcal N(A_n)$ and we should have $\operatorname{rank}|A_n|=\operatorname{rank}A_n$ as well.