Is the determinant of $\lambda I-A$ product of diagonal entries, if for every $\sigma\neq \sigma_0$, $a_{1\sigma(1)}\dots a_{n\sigma(n)}=0,$?

Question

How to prove the following assertion?

The following assertion is a piece of argument to prove Birkaff theorem(Ref. Matrix Analysis by Horn and Johnson, Chapter $8$, Lemma $8.7.1$)

Let $I\neq A$ be a doubly stochastic matrix. If for every permutation $\sigma \neq \sigma_0$, $$a_{1\sigma(1)}\dots a_{n\sigma(n)}=0,$$ where $\sigma_0$ is identity permutation, then $$\det\left(\lambda I-A\right)=\prod_{i=1}^n(\lambda-a_{ii})+\sum_{\sigma\neq \sigma_0} \operatorname{sgn}(\sigma)\prod_{i=1}^n -a_{i\sigma(i)} \tag{1}$$

The first term in the above is w.r.t the identity permutation in Laplace Expansion. But the second term doesnot seems like correct?

Is that correct statment? If so, kindly give me the arguments.

Thanks in advance.