$n \ge 2, (p,n) \ne (2,2).$ Prove that for any $v \in \mathbb{F}_p^n,$ the sequence $v, (I+M)v, (I+M+M^2)v, \dots$ has period $
Let $p$ be prime and $n \ge 2, (p,n) \ne (2,2), M \in \mathbb{M}_{n \times n}(\mathbb{F}_p).$ Prove that for any $v \in \mathbb{F}_p^n,$ the sequence $v, (I+M)v, (I+M+M^2)v, \dots$ has period $<p^n.$
My attempt: Let $M = AJA^{-1}.$ Then $(1+M+\dots+M^{r-1})v = (1+M+\dots+M^{s-1})v \Rightarrow M^r v = M^s v \Rightarrow A(J^r - J^s)A^{-1}v = 0 \Rightarrow J^r w = J^s w$ where $w = Av.$
Let $f \in \mathbb{F}_p[x]$ be the characteristic polynomial of $M.$ Let $f = qr$ where $q$ fully factors in $\mathbb{F}_p$ and $r$ is irreducible. Since $r$ is irreducible, it has no double roots. Thus, we can write $J = \text{diag}(J_1, \dots, J_k, D)$ where $D$ is a diagonal matrix, $J_i = \lambda_i I + N_i, \lambda_i \in \mathbb{F}_p,$ and $N_i$ is the matrix of ones above the diagonal of $J_i.$
Since $J_i$ has size $\le n,$ we have $N^n = 0.$ So if $p^k \ge n,$ then $J_i^{p^k} = \lambda_i I + N_i^{p^k} = \lambda_i I,$ hence $J_i^{p^k} w = \lambda_i w$ for all such $k.$
Write $D = \text{diag}(\beta_1, \dots, \beta_{\deg r})$ where the $\beta_i$ are roots of $r$ and note that $m = [\mathbb{F}(\beta_1, \dots, \beta_{\deg r}) : \mathbb{F}] \le \deg r \le \deg p = n.$ All roots lie inside $\mathbb{F}_{p^m},$ so $D^{p^m} = D.$
$J^r w = J^s w$ if and only if $J_i^r w' = J_i^s w', D^r w' = D^s w'$ where $w'$ is a portion of $w$ corresponding to $J_i$ or $D.$ Since $p^{n-1} \ge 2^{n-1} \ge n,$ we may take $r = n-1, s = n$ to ensure the first equation holds. The 2nd equation holds if $r = p^m, s = 1.$ However, I'm having trouble satisfying both equations simultaneously. What choice of $1 \le r \ne s \le p^n$ will make both equations hold, thereby proving the period is $\le s-r < p^n,$ and how could we derive or motivate the choice? If the choice depends on $v,$ that's bad news for my approach so far, but at least I've made tons of progress.
 A: It turns out a different approach is needed. As mentioned earlier, we need to rule out the case that the period is exactly $p^n.$ Suppose this is the case, and define $f_k(v) = (I+M+\dots+M^{k-1})v$ so that $\mathbb{F}_p^n = \{f_0(v), \dots, f_{p^n}(v)\}.$ If $g$ is a polynomial and $g(M)v = 0,$ then $g(M)f_k(v) = 0$ for all $k,$ hence $g(M) = 0$ identically. Note that $f_{k+1}(v) = h(f_k(v))$ where $h(v) = v+Mv,$ so $h$ has period exactly $p^n.$
Since $h(0) = v = f_1(v),$ we must have $f_{p^n}(v) = 0,$ so $(I-M)(1+M+\dots+M^{p^n - 1}) = M^{p^n} - 1$ by our earlier observation. Let $q(t)$ be the minimal polynomial of $M.$ Then $q(t+1)$ divides $(t+1)^{p^n} - 1 = t^{p^n} \mod p,$ so $q(t) = (t-1)^a$ for some $a.$ By Cayley-Hamilton, $a \le n.$ It follows $N = M-I$ is nilpotent with $N^n = N^a N^{n-a} = 0$ specifically.
Now $f_{p^{n-1}}(v) = \sum\limits_{i=0}^{p^{n-1}-1} (N+I)^i v = \sum\limits_{i=0}^{p^{n-1}-1} \sum\limits_{j \in \mathbb{Z}} \binom{i}{j} N^j  v = \sum\limits_{j \in \mathbb{Z}} \binom{p^{n-1}}{j+1} N^j v = N^{p^{n-1}-1} v$ since the other binomial coefficients vanish. If $n=2, p>2,$ then $p^{n-1} = p > n,$ and if $n \ge 3,$ then $p^{n-1} \ge 2^{n-1} > n$ as well. Thus, the remaining term is also zero and $f_{p^{n-1}}(v) = f_{p^n}(v) = 0,$ contradiction.
