Let $A$ be an $N \times N$ matrix with nonnegative entries such that for a certain power $A^n$ all entries are positive. Consider the unit simplex $\sigma = \{ (x_1, ..., x_N) : x_i \ge 0, x_1 + ... + x_N = 1\}$ and let $T : \sigma \to \sigma$ be a map defined by $$T x = \frac{A x}{\|Ax\|}.$$ The set $\sigma_0 = \bigcap_{n=0}^\infty T^n \sigma \subset Int (\sigma)$ is closed, convex and $T \sigma_0 = \sigma_0$. Let $x \in \sigma_0 \subset T^n \sigma$. Then $x$ is a convex linear combination of extreme points of $T^n \sigma$, but all extreme points of $T^n \sigma$ are among the images of the vertices $e_1, ..., e_N$ of $\sigma$. Thus $$x = \sum_{i=1}^N \lambda_i^{(n)} T^n e_i,$$ where $\lambda_i^{(n)} \ge 0$ and $\sum_{i=1}^N \lambda_i^{(n)} = 1$.
Can someone explain me why a sequence $n_k \to \infty$ can be found such that $T^{n_k} e_i$ and $\lambda_i^{(n_k)}$ converge for all $i = 1, ..., N$?
Because the sequences $(T^ne_i)_n$ and $(\lambda_i^n)_n$ are bounded in $\mathbb R^n$ and $\mathbb R$. And there are only a finite number of such sequences. Just use a subsequence argument $2N$-times.