Let y be an nth root of 1, hence x-y divides xn-1. We have found one of the prime factors of xn-1 in K[x]. Yet y2 also satisfies xn = 1, hence x-y2 is another prime factor of our polynomial. This holds for all powers of y up to n.
Note that yi cannot equal yj, else yi-j = 1, and y is actually a lower root of 1. If y is a primitive nth root of 1, such that n is the least positive exponent satisfying yn = 1, then all powers of y up to n are distinct. This means all binomials x-yj are distinct. We have found the n prime factors that divide xn-1. This is the complete factorization for the polynomial, and the following must hold.
xn-1 = u × (x-y) × (x-y2) × (x-y3) × … × (x-1)
Here u is some unit, but by looking at the coefficient on xn, the unit must be 1, so we don't have to worry about that.
Expand the product and note that the coefficient on xn-1 must be 0. Therefore the sum of all nth roots of 1 is 0. The coefficient on xn-2 is also 0, hence the sum of the pairwise products of the nth roots of 1 is also 0. This continues all the way down the line, until the product of the roots of 1 yields -1.
Consider the product of the factors s-tyj, where s and t are arbitrary symbols, and j runs from 1 to n. As shown above, all the coefficients, other than the first and last, drop to 0. Therefore the product is sn-tn.
If n is odd, replace t with -t. Thus the product over s+tyj yields sn+tn.
Adjoining y to K brings in all the powers of y, all the roots of xn-1. This makes K(y) a splitting field. The extension is normal. And since K is perfect, the extension is also separable. A normal separable extension is galois, so K(y) is galois.
An automorphism necessarily moves y to another primitive nth root of 1. Thus y maps to yj, where j and n are relatively prime. Once y is mapped, the entire automorphism is established. Therefore the galois group of this extension is the group of units mod n, denoted Zn*.
Actually there is a catch. Some of the higher powers of y could lie in K, and some of the automorphisms might move these values about. In other words, some of the automorphisms might not fix K. This is not a problem when K = Q. Every automorphism maps 1 to 1 and -1 to -1, and all other nth roots are complex, and lie outside of Q. All the automorphisms fix Q, and the galois group is indeed Zn*.
When K is a finite field, some of the powers of y might lie in K. These move about under certain automorphisms, hence the galois group becomes a subgroup of Zn*.
Illustrate with the 24th roots of 1 over Z5. Let F be the finite field of order 25. Every nonzero element of F is now a 24th root of 1. If y is a primitive element of F, then F = Z5(y). However, the galois group of this extension is not isomorphic to the group of units mod 24. Some of these automorphisms swap 2 and 3, and mess with the base field K. As it turns out, the extension F/K has dimension 2, and its galois group is Z2. This is a subgroup of Z24*.
So far we have assumed that a primitive root exists. This is certainly the case when K is the rationals. The unit circle in the complex plane can always be cut into n equal arcs. But what about finite fields?
Let n = p×t. The polynomial xn-1 is now xt-1 raised to the p power. The roots of xt-1 are adjoined p times over. This doesn't make much sense, so assume p does not divide n.
If F/K is a finite field extension, F has order pj for some exponent j. Remember that F* is cyclic. Now y is contained in F iff n divides pj-1. We are looking for an exponent j such that pj is 1 mod n. Since p is coprime to n it is a unit mod n. In the group of units n*, there is always some exponent j such that pj = 1 mod n. There is always some finite field extension F containing the nth roots of 1.