Let $H$ be a $\mathbb R$-Hilbert space, $A\in\mathfrak L(H)$ be compact and $(\sigma_n)_{n\in\mathbb N}$ be a nonincreasing enumeration of the singular values of $A$ counted with multiplicity. By definition, the singular values of $A$ are the eigenvalues of the square-root $$|A|:=\sqrt{A^\ast A}.$$ Moreover, by the Courant-rayleigh minimax principle, $$\sigma_k=\max_{\substack{U\le(\ker A)^\perp\\\dim U=k}}\min_{\substack{x\in U\\\left\|x\right\|_H=1}}\langle|A|x,x\rangle_H\tag1\;\;\;\text{for all }k\in\mathbb N.$$ In particular, $$\sigma_1=\left\||A|\right\|_{\mathfrak L(H)}=\left\|A\right\|_{\mathfrak L(H)}\tag2.$$
Now, I wondered whether we are able to relate the $\sigma_k$ to the tensor product $A^{\otimes k}$. By definition, if $L_1\in\mathfrak L(H)$, $L_1\otimes L_2$ is the linearization of $$H^2\ni(x_1,x_2)\mapsto L_1x_1\otimes L_2x_2.\tag3$$ The question is, with which norm (or inner product) we need to endow the space of this operators.
I would like to find a relation of the kind $$\left\|A^{\otimes k}\right\|=\sigma_1\cdots\sigma_k\tag4.$$
Obviously, if $(e_1,\lambda_1),\ldots,(e_n,\lambda_n)$ are eigenpairs of $A$, then $$A^{\otimes n}\left(\bigotimes_{i=1}^ne_i\right)=\prod_{i=1}^n\lambda_i\bigotimes_{i=1}^ne_i\tag5$$ and hence $\left(\bigotimes_{i=1}^ne_i,\prod_{i=1}^n\lambda_i\right)$ is an eigenpair of $A^{\otimes n}$. The only problem now is that we cannot apply the argumentation above again, unless $A^{\otimes n}$ is again an operator on a Hilbert space.
If I remember correctly, endowing the tensor product of operators with the projective norm and building the completion yields the space of trace-class operators (which is not a Hilbert space), but building the completion with respect to the Hilbert-Schmidt norm yields the space of Hilbert-Schmidt operators (which is a Hilbert space). So, I guess the latter is what we need to do.
EDIT: Actually, what I'm trying to find is an analogue of Proposition 3.2.7 in Random Dynamical Systems:
You seem to be assuming that $A$ is compact.
You cannot expect $(4)$ with $k>1$. It is trivial to construct nonzero $A$ with $\sigma_2=0$. Then, for any norm, you would have $\|A^{\otimes k}\|=0$, so $A^{\otimes k}=0$.
If you use the operator norm, you have $$ \|A^{\otimes k}\|=\|A\|^k=\sigma_1^k. $$