The Order of an Especial Element in the Binary Finite Field Consider $G=GF(2^{2n})$.  Let $r\in G$ such that $ord(r)=2^n-1$. We can show that  for $k=1$ and  $k \nmid 2^n-1$, we have
$$
p_k(1,r^1,\cdots,r^{2^n-2})=\sum_{i=0}^{2^n-2}r^{ik}=0
$$
Also for $k\mid 2^n-1$ we have $p_k(1,r^1,\cdots,r^{2^n-2})=1$. Suppose that $s\in G$ be a non-zero element such that $s\neq r^i$ for $0\leq i \leq 2^n-2$ or in other words the order of $s$ dose not be $2^n-1$. 
My question: Consider the following element 
$$
t=s\,(s+1)(s+r)\cdots \, (s+r^{2^n-2})
$$
How to prove that $t=r^k$ for some $0\leq k \leq 2^n-2$. 
My try: In fact, we have to show that the order of $t$ is $2^n-1$. Due to 
 $2^{2n}-1=(2^n-1)\,(2^n+1)$, it is sufficient to prove that  $t^{2^n-1}=1$. From the Newton's identities, we can write $t$ as follows
$$
t=s^{2^n}+e_1\, s^{2^n-1}++e_2\, s^{2^n-2}+\cdots + e_{2^n-1}\, s
$$
From the relation between $p_k$ and $e_k$ by Newton's identities, we obtain $e_i=0$ for $1\leq i \leq 2^n-2$ and $e_{2^n-1}=1$ which results that 
$$
t=s^{2^n}+s=s(s^{2^n-1}+1)\, \Rightarrow \, t^{2^n-1}=s^{2^n-1}(s^{2^n-1}+1)^{2^n-1}
$$
In addition, by Lucas theorem we obtain ${{2^n-1}\choose j}=1$ for $1\leq j \leq 2^n-1$, since the binary form of $2^n-1$ is $(11\cdots1)$. So we get
$$
t^{2^n-1}=s^{(2^n-1)(2^n)}+s^{(2^n-1)(2^n-1)}+\cdots + s^{(2^n-1)(2)}+s^{(2^n-1)}
$$
Therefore, we can conclude that( we add $1+1$ to the above equation)
$$
t^{2^n-1}=\frac{s^{2^{2n}-1}+1}{s^{2^n-1}+1}+1=\frac{1+1}{s^{2^n-1}+1}+1=1.
$$
Is my proof correct and I appreciate to suggest an easy proof.
Thanks for any suggestions. 
Edit: By the nice answer of @Jyrki Lahtonen, the tag is improved. 
 A: Your proof looks ok to me. 
I might go about this task as follows (just the first idea that occured to me). Let $F=GF(2^n)$ be the prescribed subfield of $G$. It is well known (from a standard proof of uniqueness of a finite field of a given cardinality) that the elements of $F$ are exactly the zeros of $P(x)=x^{2^n}-x$, all simple. In other words
$$P(x)=x^{2^n}-x=\prod_{z\in F}(x-z).$$
Below I will replace all those minus signs with plus signs because we are in characteristic two!
Your element $r$ is a generator of the multiplicative group $F^*$. Therefore we can rewrite the above factorization as
$$
P(x)=x^{2^n}+x=x\prod_{i=0}^{2^n-2}(x+r^i).
$$
From this we see immediately that
$$
t=s\prod_{i=0}^{2^n-2}(s+r^i)=P(s)=s^{2^n}+s.
$$
No need for Newton's identities here!
Because $s\notin F$ we know that $t=P(s)\neq0$. On the other hand $s\in G$, so
$s^{2^{2n}}=s$. Therefore
$$
P(t)=t^{2^n}+t=(s^{2^n}+s)^{2^n}+(s^{2^n}+s)=s^{2^{2n}}+s^{2^n}+s^{2^n}+s=0.
$$
So we have shown that $t$ is a non-zero element of the subfield $F$, i.e., a power of $r$.

Basically this was about the Galois theory of the extension $G/F$. Here the only non-trivial automorphism if $\sigma(z)=z^{2^n}$, and this is of order two. In this case $P(z)$ is the so called relative trace $P(z)=\sigma(z)+z=tr^G_F(z)$.
