suppose n is odd.

then (-1)^n+1=0 so x+1 divides 1+x^n=1+0=1 by factor theorem

so there is an element y such that (1+x)(y)=1

what if n i even?

so after you have done this you can use long division for each aswer to give you the unique y needed and you see that the elements are both of the form you wrote them in and is composed of the identity element and x, and so y is in the ring with identity.