Is this stochastic matrix has a steady state equivalent to Perron Frobenius theorem？

This is the capture from Lay-linear algebra. I am trying to find the proof under this specific statement, but I cannot find it online. Rather, I've seen Perron Frobenius theorem which is very close to this(I am not sure), something about maximal eigenvalue is $1$ involving contraction mapping.

I am wondering if anyone could help me out to prove this specific statement if they are not equivalent, or redirect me to the proof of this specific statement.

From a stochastic matrix $P^{n \times n}$ we can compute the distribution of the Markov chain in $X_{n+1}$ from induction computing the distribution of $X_{n}$ using the law of total probability.

Observe that the space $X$ is finite, with $\mathbb{X}=\{v_{1},v_{2},\dots,v_{n}\}$ we can describe the probability mass function as a vector :

$$\pi = (\pi(v_{1},\pi(v_{2},\dots,\pi(v_{n})))) $$

or $$\pi_{n+1} = \pi_{n}P$$ where in the rhs is the regular matrix multiplication.

Defining the power $$P^{n} =\{p^{(n)}(x,y) \}_{x,y \in \mathbb{X}}$$ of a stochastic matrix $P$ for $n \in \mathbb{N}_{0}$ and using the Fubinni - Tonelli theorem we have:

$P^{0}=I$ and in general $$P^{n}=P^{n-1}P $$ for $n \in \mathbb{N}$

Diagonilizing the matrix $P$ we will obtain the eigenvalue decomposition looking like :

$$P^{n} = \begin{pmatrix} 1 & u_{1}\\ 1 & u_{2} \end{pmatrix}\begin{pmatrix} 1 & 0\\ 0 & \lambda^{n} \end{pmatrix}\begin{pmatrix} 1 & u_{1}\\ 1 & u_{2} \end{pmatrix}^{-1}$$

taking the limit where $\lim_{n \to \infty} \lambda^{n}$ we can reach the limiting distribution.

Doing so we have : $$\pi_{n} = \pi_{0}P^{n}$$. Observe at last that the initial vector distribution does not play any role at all in the limiting distribution.The steady state is been obtained by taking the limit in the second or in the $n^{th}$ eigenvalue.

In general $$P^{n} = Q\Lambda^{n}Q^{-1}$$ and taking the $\lim_{n \to \infty}$