Proving 1+x is a Unit in a Ring for x^n=0

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Homework Statement



Let R be a ring and x in R such that x^n=0 for some n show that 1 + x is a unit.

I know then that x is a zero divisor and I need to find y such that y(1+x) = 1.
I can see in examples that this works and I can prove it for Z mod n. I can't figure out how to prove it for any ring. Please help
Thanks
 
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If there exist n such that xn, there exist a smallest such n. Assume, without loss of generality that n is the smallest number such that xn= 0. If n= 1, then x= 0, x+1= 1 which is a unit. If n> 1, then xn-1 is not 0. Let u= xn-1. Then u(1+ x)= xn-1+ xn= xn-1= u. Does that lead anywhere? In particular is u a unit?
 
You probably know that you can write 1/(1+x) formally as a power series, 1-x+x^2-x^3+... If x^n=0, that series terminates. Can you show that it's true that (1+x)*(1-x+x^2-...x^(n-1))=1?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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