Say the rank of a matrix drops for all powers up to $n$ $$\text{rank}A^n< \text{rank} A^{n-1} < \ldots <\text{rank} A$$
Now suddenly $\text{rank} A^n = \text{rank} A^{n+1}$. Does the following implication hold true then $$ \text{rank}A^n = \text{rank}A^{n+k} \,, \forall \, k \in \mathbb{N}$$
I am asking this to know when to stop looking for the next vektor in a Jordan chain.
On
Rank is the dimension of the column space (i.e. the image of the linear transformation).
It is always true that $Col(A^{m+1}) \subseteq Col(A^m)$, or that $im \ A^{m+1} \subseteq im \ A^m $ (The more general fact that $Col(AB)\subseteq Col(A)$ is not difficult to prove.)
So let's consider sequence of images $V \supseteq im\ A \supseteq im \ A^2 \supseteq...$
Initially, $A $ reduces (or maintains, if it is an isomorphism) the dimension of $V$ to that of its image, then $A^2$ further reduces the dimension to $im \ A^2$ via $A$ restricted to $im \ A $ (since after the first application only $im \ A$ is nonzero, and linear maps send zero to zero). So when the sequence stabilises, $im \ A^m = im \ A^{m+1} $, this essentially means that $A$ restricted to $im \ A^m$ is an isomorphism. Since any power of an isomorphism is an isomorphism, no further iterations of $A$ will reduce the dimension.
Note also that by the rank-nullity theorem, as the rank decreases the kernel increases, and the two sequences stabilise at the same step.
If we are allowed to use Jordan canonical form, the answer is obviously yes, as $n$ must be the size of the maximal Jordan block with $0$ on the diagonal.
To show this directly, we have $\text{rank}A^n=\text{rank}A^{n+1}$ is equivalent to $\text{im}(A^{n+1})=\text{im}(A^n)$, in other words, $A$ is a bijection from $\text{im}(A^n)$ to itself. Now it follows that $A^{k}$ is a bijection from $\text{im}(A^n)$ to itself.