Question about a proof of the Perron Frobenius Theorem

Question

I am going through a proof of the Perron-Frobenius theorem that I found online, but I'm having trouble understanding some of the steps. Here is the relevant info:

Let $T$ be a primitive row-stochastic matrix. (By "primitive" we mean there exists an integer $k$ such that all entries of $T^k$ are strictly positive). I have been working through the proofs of the following statements:

1. $1$ is an eigenvalue of $T$.
2. If $\lambda$ is an eigenvalue of $T$ then $|\lambda| \leq 1$. (And hence $\lambda_{\max} = 1$ is the largest eigenvalue.)
3. The algebraic multiplicity of $\lambda_{\max} = 1$ is one.

My current status:

1. This follows immediately from the fact that $T(1 \; 1 \; \cdots 1)^{\top} = (1 \; 1 \; \cdots 1)^{\top}$, since $T$ is row-stochastic.
2. It can be shown that $ |\lambda | \|\vec{x} \|_{1} = \|\lambda \vec{x} \|_{1} = \|T \vec{x} \|_{1} \leq \|x\|_{1}$. The last inequality is obtained by some basic algebra and the Triangle Inequality. The result follows.
3. This is where I am having difficulty. The proof I am reading through obtains the result using the following Lemma: "Let $T$ be a (square) matrix and let $\Lambda$ be a diagonal matrix of the same size with entries $\lambda_1,\lambda_2,\ldots,\lambda_n$ along the diagonal. Expanding $\det(\Lambda - T)$ along the $i$-th row shows that $$\frac{\partial}{\partial \lambda_i} \det(\Lambda - T) = \det(\Lambda_i - T_i),$$ where the subscript $i$ means the matrix obtained by eliminating the $i$-th row and $i$-th column from each matrix." So far, this makes sense to me...The proof continues: "Setting $\lambda_i = \lambda$ and applying the Chain Rule from calculus, we get $$\frac{d}{d \lambda} \det(\lambda I - T) = \sum_i \det(\lambda I - T_{i}) \hspace{2cm} (*)$$

Each of the matrices $\lambda_{\max}I - T_i$ has strictly positive determinant by what we just proved. This shows that the derivative of the characteristic polynomial of $T$ is not zero at $\lambda_{\max}$, and therefore the algebraic multiplicity and hence the geometric multiplicity of $\lambda_{\max}$ is one."

I don't follow how (*) was obtained...How exactly is the Chain Rule being applied here? And where did the summation come from? And how can we say that $\det(\lambda_{\max}I-T_i) >0$ for all $i$? The author says this is by "what we just proved"...Where was that proved?

Probably I am making some very silly oversight, but any insights or help would be greatly appreciated!