$\Lambda^k(A)$ is a trace class operator

Question

Let $A$ be a trace class operator. I am trying to understand the proof of ($\|\cdot\|_1$ means the trace class norm)

$$\|\Lambda^k(A)\|_1\leq \frac{\|A\|_1^k}{k!}$$

I have a proof but I think there is a mistake in it.

In the proof is the followig: $|\Lambda^k(A)| =\Lambda^k(|A|)$ with eigenvalues $\mu_{i_1}(A)\cdots \mu_{i_k}(A)$ for $i_1<\cdots <i_k$ so $$\sum\limits_{i_1<\cdots< i_k} \mu_{i_1}(A)\cdots \mu_{i_k}(A)=\frac{1}{k!}\sum\limits_{i_1<\cdots< i_k} \mu_{i_1}(A)\cdots \mu_{i_k}(A)$$

The equality doesn't make sense for me.

Answer 1

The second summation should probably be without $i_1<...<i_k$, so that would be $k!$ duplicates.