i probably wasn't very clear in my previous post.

in your proof you said $\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x-1)^{2^n},$ which is clearly not correct even if we assume the ring has 1. the correct one is

$\displaystyle (x^{2}-x)^{2^n+1} = x(x-1)(x^2-x)^{2^n}.$ anyway, since there is nothing to prove if n = 0, we may assume that $\displaystyle n > 0.$ as you mentioned $\displaystyle 2 \mid \binom{2^n}{k},$ for all $\displaystyle 0 < k < 2^n.$ therefore, since $\displaystyle 2r=0$ for all

$\displaystyle r \in R,$ we have:

$\displaystyle x^2 - x = (x^2 - x)^{2^n + 1} = (x^2 - x)(x^2 - x)^{2^n}=(x^2 - x)(x^{2^{n+1}} + x^{2^n})$

$\displaystyle =x^{2^{n+1} + 2} + x^{2^n + 2} - x^{2^{n+1}+1} - x^{2^n + 1}=x^2 +x^2 - x - x = 2x^2-2x = 0.$