Does one of the orthogonal groups $O^\pm(2n,2)$ always have an element of order $2n+1$ For arbitrary $n\ge 1$, is it true that (at least) one of the orthogonal groups $O^\pm(2n,2)$, consisting of $2n\times 2n$ matrices over the field of order $2$, has an element of order $2n+1$?
A quick check in MAGMA shows that it is true for $n\le 7$ (code still running for $n=8$). There is not an obvious pattern in the matrices found, but there are many solutions in each case, so some nice construction may be possible. 
EDIT: Reducing to $2n+1$ prime
We can reduce to $2n+1$ prime as follows. This means that it is sufficient to prove that $2n+1$ divides the order of one of $O^\pm(2n,2)$ when $2n+1$ is prime. A quick check shows that this is true for primes $p<1000$.
Suppose the above is true when $2n+1$ is prime and $2n+1$ is the smallest non-prime for which the above does not hold. Then $2n+1=(2u+1)(2v+1)$ for some $u,v>1$ and $O^\epsilon(2u,2)$ has an element $M$ of order $2u+1$ for some $\epsilon\in\{\pm\}$. One can check then that the following is an element of $O^{\epsilon'}(2n,2)$ (in the appropriate basis for some $\epsilon'\in\{\pm\}$) and has order $2n+1$ contrary to assumption:
$$\left(\begin{matrix}
\mathbf{0}_{2u} & M & \mathbf{0}_{2u} & \cdots & \cdots & \cdots & \mathbf{0}_{2u} & \mathbf{0}_{2v} \\
\mathbf{0}_{2u} & \mathbf{0}_{2u} & \mathbf{1}_{2u} & \mathbf{0}_{2u} & \cdots & \cdots & \mathbf{0}_{2u} & \mathbf{0}_{2v} \\
\mathbf{0}_{2u} & \cdots & \mathbf{0}_{2u} & \mathbf{1}_{2u} & \mathbf{0}_{2u} & \cdots & \mathbf{0}_{2u} & \mathbf{0}_{2v} \\
\vdots & \ddots & \ddots & \ddots & \ddots & \ddots & \vdots & \vdots \\
\mathbf{0}_{2u} & \cdots & \cdots & \cdots & \cdots & \mathbf{0}_{2u} & \mathbf{1}_{2u} & \mathbf{0}_{2v} \\
\mathbf{1}_{2u} & \mathbf{0}_{2u} & \cdots & \cdots & \cdots & \cdots & \mathbf{0}_{2u} & \mathbf{0}_{2v} \\
\mathbf{0}_{2u} & \cdots & \cdots & \cdots & \cdots & \cdots & \mathbf{0}_{2u} & \mathbf{1}_{2v}\\
\end{matrix}\right)$$
 A: The symmetric group $S_{2n+1}$ embeds into ${\rm SO^+}(2n,2)$ when $n \equiv 0,3 \bmod 4$ and into ${\rm SO^-}(2n,2)$ when $n \equiv 1,2 \bmod 4$, via the deleted permutation module.
This is proved, for example, on page 187 of Kleidman and Liebeck's book "The Subgroup Structure of the Finite Classical Groups" where the quadratic form that is preserved is identified (well, actually, they do it for $A_{2n+1}$, but the proof works for $S_{2n+1}$).
A: Edit: The answer immediately under is completely false and reflects a dreadful understanding of the problem. The part underneath is an actual solution attempt. 


False solution: 
There is a matrix $M \in SO(2n)$ with order $2n+1$, so consider $\begin{bmatrix} M & O \\ O & I_2\end{bmatrix}$?

 

Actual partial solution (not for all $n$; addresses only one of the orthogonal groups):
First, let $d \geq 3$ be an odd integer, assume that $d|2^{\nu_d}+1$ for some minimal $\nu_d > 0$, then $\omega_d=2\nu_d$ is the multiplicative order of $2$ mod $d$.
Consider a primitive $d$-th root of unit $\zeta_d$ in a finite (of dimension $\omega_d$) extension $F$ of $\mathbb{F}_2$. 
Let $T$ denote the trace from $F$ to $\mathbb{F}_2$. Note that for every $u,v \in F$, $(uv^{2^{\nu_d}})^{2^{\nu_d}}=u^{2^{\nu_d}}v$, thus $T(uv^{2^{\nu_d}})=T(u^{2^{\nu_d}}v)$. 
Denote $\beta_d(u,v)=T(uv^{2^{\nu_d}})$, it is a nondegenerate $\mathbb{F}_2$-bilinear symmetric form on $F$. 
Note furthermore that $\zeta_d^{1+2^{\nu_d}}=1$, by definition of $\nu_d$. 
Therefore, the endomorphism $f_d : u \in F \longmapsto \zeta_d u \in F$, has order $d$, and is orthogonal for $\beta_d$. 
Now, let $n \geq 1$ be an integer such that mod $2n+1$, $-1$ is a power of $2$. Then for all $1 < d \leq 2n+1$, $d|2n+1$, $f_d$ and $\beta_d$ are defined (for suitable, if arbitrary, choices of roots of unity). 
Let, for all such $d$, $r_d=\frac{\varphi(d)}{\omega_d}$, and consider $f$ (resp, $\beta$) to be the direct sum over all $d$ of $r_d$ times $f_d$ (resp. $\beta_d$). 
Then $\beta$ is a nondegenerate $\mathbb{F}_2$-bilinear symmetric form, defined over a $\mathbb{F}_2$-vector space $V$ with dimension $2n$, and $f$ is an orthogonal endomorphism of $(V,\beta)$. 
Furthermore, since for all $d$, $d |2n+1$, thus $f_d^{2n+1}=id$, so $f^{2n+1}=id$. On the other hand, if $f^r=id$, then $f_{2n+1}^r=id$ thus $2n+1|r$, so $f$ has multiplicative order $2n+1$. 


 
 
Edit: similar approach, but it works for more $n$, eg $2n+1=15$.
Let $D$ denote the set of divisors of $2n+1$ distinct from $1$, $P$ the set of elements of $D$ that are primes or prime powers, $Q$ the set of elements of $P$ that are primes. 
Assume that for all $p \in Q$, the multiplicative order of $2$ mod $p$ is even (that is, $-1$ is a power of $2$). Then, for all $d \in P$ (with LTE), $-1$ is a power of $2$ mod $d$, so we can consider $\omega_d$, $\beta_d$, $f_d$ as above.
Since the sum of the $\varphi(d)$, $d \in D$, is $2n$, and since $\varphi$ is multiplicative, there exists positive integers $r_d$, $d \in P$ such that $\sum_{d \in P}{r_d\varphi(d)}=2n$. 
So we consider $f$ (resp, $\beta$) to be the direct sum (over $d \in D$) of $r_d\frac{\varphi(d)}{\omega_d}$ functions $f_d$, (resp. $\beta_d$), and we conclude as above. (The order is slightly less obvious, but works nonetheless)
