<s>[INST] $(x^n)^2=x^n$ for all $n\geq 1$. $x$ is idemotent ring where $x^2= x$ I am asked to prove the above 
I have been trying a lot but my teacher did not like they way I solved it,he did not even tell me what is wrong with it.i was thinking if $n=3$ then $x^{2n}$ will be equal to $x^6$ is equal to $$x\cdot x\cdot x\cdot x\cdot x\cdot x=x^6=x^2\cdot x^2\cdot x^2$$
But we know that $x=x^2$
So $x^{2n}=x\cdot x\cdot x = x^3=x^n$
"............sorry can't answer my question now but is this what u mean
(x^n)^2 = x^n for all n>= 1
now for n = 1, we have 
x^2 = x which is true for x a member of r
for n+1 we need to show
(x^n+1)^2 = x^n+1
(x^n+1)^2 = (x^n * x)(x^n * x)
(x^n+1)^2 = (x^n)^2 * x^2
(x^n+1)^2 = x^n * x
(x^n+1)^2 = x^n+1 
 [/INST] Hint: $(x^n)^2=x^{2n}=(x^2)^n$.
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