Characteristic $2$ ring $A$ with $x^{2^k +1}=x$ for some $k$ for every $x \in A$. Show that $x^2 = x$. 
Let $A$ be a ring of characteristic $2$ with $x^{2^k +1}=x$ for some $k$ for every $x \in A$ (i.e. $k$ is specific to $x$). Show that $x^2 = x$ for every $x \in A$.

I have been fiddling about for a while now but have not had much luck. I have to tried to use the fact $A$ is characteristic 2 e.g. by looking at different bracket expansions, however since we don't know any properties about $A$ (e.g. domain?) I can't seem to equate these expressions to anything well known or nice. Perhaps I am missing an obvious short rearrangement trick.
Any hints/ideas would be appreciated.
 A: Hint: to prove $x^2=x$ you only need to use the condition twice:


*

*there exists $k\geq 1$ with $x=x^{2^k+1}$

*there exists $b>1$ such that $(1+x)^{2^b+1}=1+x$

A: We start with a lemma: By Theorem 2 of Binomial coefficients modulo a prime (Fine, 1947), we have that if the binary expansion of the positive integer $N$ has $m$ ones on it, then the number of odd entries of the $N$-th row of Pascal's Triangle is $2^m$. In particular, if $N=2^k$ then $m=1$ and $2^m=2$, so the only odd entries are the $1$s from the extremes. Similarly, if $N=2^k+1$ (which is odd) then $m=2$ and $2^m=4$, so the only odd entries are $1, N, N, 1$ at the extremes.
Now let $k$ and $n$ be such that $x^{2^k+1}=x$ and $(x+1)^{2^n+1}=x+1$.
By the lemma we have $$x+1=(x+1)^{2^n+1} = x^{2^n+1}+x^{2^n}+x+1,$$ which implies $x^{2^n+1}+x^{2^n}=0$, or $$x^{2^n+1}=x^{2^n}.$$
By the lemma we also have $$(x+1)^{2^k+1} = x^{2^k+1}+x^{2^k}+x+1 = x+x^{2^k}+x+1 = x^{2^k}+1 = (x+1)^{2^k},$$ the last step again by our lemma.
We have to put these results together. We will use that, for a big enough power of $x$, the chain of powers of $x$ stabilizes, while for a bigger power $M$ of $x$, the "cycle is closed" and we get $x^M=x$.
Suppose $n<k$, so that $2^n<2^k+1$. Since $x^{2^n}=x^{2^n+1}$, we infer $x^{2^n}=x^s$ for every $s>2^n$, hence
$$x=x^{2^k+1}=x^{2^n}=x^{2\cdot 2^n}=(x^{2^n})^2=x^2.$$
If on the contrary $n\geq k$, then as $(x+1)^{2^k}=(x+1)^{2^k+1}$, we infer $(x+1)^{2^k}=(x+1)^s$ for every $s>2^k$, so that
$$x+1=(x+1)^{2^n+1}=(x+1)^{2^k}=(x+1)^{2\cdot 2^k} = ((x+1)^{2^k})^2 = (x+1)^2 = x^2+1,$$
which clearly implies $x=x^2$.
