Finding order of a matrix I am trying to find the order, i.e. smallest $k\in \mathbb{N}, k>1$ s.t. $A^k=I$, of the following matrix where arithmetic is over $\mathbb{Z}_2$:
$$A=\begin{bmatrix}0&1&0&\cdots 0\\
0&0&1&\cdots 0\\
\vdots\\
0&0&0&\cdots 1 \\
1&1&1&\cdots1
 \end{bmatrix}_{n\times n}$$
I've looked at a few $n$ and looks like the order is $n+1$ but I am not sure how to prove that? Any ideas? I see that with increasing power by one shifts original rows of $A$ up by one and adds rows corresponding to the identity matrix. For example, for $n=4$:
$$A^2=\begin{bmatrix}0&0&1&0\\
0&0&0&1\\
1&1&1&1 \\
1&0&0&0
 \end{bmatrix},A^3=\begin{bmatrix}0&0&0&1\\
1&1&1&1\\
1&0&0&0 \\
0&1&0&0
 \end{bmatrix},A^4=\begin{bmatrix}1&1&1&1\\
1&0&0&0\\
0&1&0&0 \\
0&0&1&0
 \end{bmatrix},A^5=\begin{bmatrix}1&0&0&0\\
0&1&0&0\\
0&0&1&0 \\
0&0&0&1
 \end{bmatrix} $$
 A: Over the finite field $\Bbb F_2$ the order of $A\in GL_n(\Bbb F_2)$ is $n+1$. Indeed, the row $(1,1\cdots ,1)$ is rising up one level after each multiplication with $A$. So $A^{n+1}=I$ and $A^n\neq I$. Formally, let $(e_1,\ldots ,e_n)$ be a basis. Then compute $A^{n+1}(e_i)$ for each $i=1,\ldots ,n$.
Over a field of characteristic zero, $A$ has infinite order.
A: for similar but more linear algebra oriented proof:
i.) $A$ is a Companion matrix of the polynomial $x^n+x^{n-1}+...+x+1$
ii.) By Cayley Hamilton $\mathbf 0 = A^n+A^{n-1}+...+A+I$
iii.) over any field, the minimal polynomial of a Companion matrix is the characteristic polynomial.  This implies the order of $A$ is $\gt n$.
iv.) 'Subtracting' (or adding since the field has characteristic two) $A^n$ from each side in (ii.), and then multiplying each side by $A$, we get
$A^{n+1}=A\big(A^n\big)= A\big(A^{n-1}+...+A+I\big)= A^{n}+A^{n-1}...+A=I$
where the final equality comes from adding $\mathbf 0 =\big(A^n+A^{n-1}+...+A+I\big) $ to $A^{n}+A^{n-1}...+A$
