Let $K$ be a field of characteristic distinct to $2$ and $n$ be a natural. I want to know if my proof of the existence of $A\in \text{GL}_n(K)$ of order $n+1$ is correct:
Consider $A$ to be the companion matrix of the polynomial $p=x^n+\cdots+x+1$. First note that: $$A^n+\cdots+A+I=0\implies A^{n+1}=\underbrace{-A^n-\cdots-A-I}_{=0}+I=I.$$ Therefore, $\text{ord}A\mid n+1$. Now, suppose there exists $0<i\leq n$ such that $A^i=I$. Then, $$\text{polmin}A=x^n+\cdots+1\mid x^n+\cdots+x^{i+1}+x^{i-1}+\cdots+1+1,$$ which is imposible because the polynomial on the right is not zero, has degree less or equal to $n$ and is different to $p$.
The case of a field of characteristic $2$ and $n=1$ is different. In $\text{GL}_1(\mathbb{Z}_2)=\{\overline{1}\}$ there is no matrix of order $2$. I think this is the only case that fails. Any help would be appreciated.