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.