this is the only proof i know:

lemma 1: every polynomial f(x) in F[x], for a field F has at most n roots.

proof: fields are euclidean domains with d function deg(f). the result follows by induction on the degree of a polynomial and the division algorithm. (note: a purely analytic argument using the continuity and differentiability of polynomials and rolle's theorem and the intermediate value theorem is also possible).

lemma 2: the polynomial x^{n}- 1 has n roots in the field of complex numbers

proof: by demoivre's theorem, the complex numbers z_{k}= cos(2πk/n) + i sin(2πk/n) for k = 0,1,2,...,n-1 are all roots of x^{n}- 1. by lemma 1, these are all the roots of x^{n}- 1.

lemma 3: if a > 0 is a real number, the polynomial x^{n}- a has at least one real root.

proof: since x^{n}- a is a real polynomial, it's complex roots occur in conjugate pairs: that is if z^{n}= a, then (z^{n})* = a* = a so z and z* are both n-th (complex) roots of a.

if n is odd, we must have z = z* for some n-the root z of a, which is therefore a real root of x^{n}- a.

on the other hand, if n is even, then for some positive integer N, we have N^{2}- a > 0, so that x^{n}- a is positive on (-∞,N) and (N,∞), since 0^{2}- a = -a < 0, by the intermediate value theorem, x^{n}- a must have a root in (N,0) and another root in (0,N).

(alternatively: if you are willing to accept the theorem as true for n = 2, then if n is even let y = x^{n/2}, and let √a be the square root (the unique positive one that exists in the case of n = 2, by assumption), so we have:

x^{n}- a = (y - √a)(y + √a). thus x^{n}-a has a solution if x^{n/2}- √a has a solution, if n/2 is odd, we are done. if not, repeat. if n is a power of 2, eventually we will reach the case where we have x^{2}- b

(where b is some square root of a square root of a..(depending on how many times we have to do this)..square root of a)). if n has any other factor, eventually we'll factor out all the powers of 2, and terminate with an odd case).

lemma 4: let u be a real root of x^{n}- a, and let w = cos(2π/n) + i sin(2π/n). then {u,uw,uw^{2},...,uw^{n-1}} are the n roots of x^{n}- a.

proof: follows by lemma 1, lemma 2 and lemma 3 (lemma 3 for the existence of u, lemma 1 to show these are all the possible roots, lemma 2 to show these actually are roots).

now all we have to do is show we can take u > 0, and that there is always exactly one such u.

suppose uw^{k}is a real root of x^{n}- 1. then:

(uw^{k})* = uw^{k}that is:

u*(w^{k})* = uw^{k}, hence:

(w^{k})* = w^{k}, so w^{k}is real. the numbers w^{k}all lie on the unit circle |z| = 1, which intersects the real line in just two places: 1 (= (1,0) = 1+i0) and -1 (= (-1,0) = -1+i0).

thus 2πk/n = π or 2πk/n = 0. in the first case, we have k = n/2, and since the k's are integers, this can only happen when n is even. in this case we get two roots: u and uw^{n/2}= -u.

exactly one of these is positive. if n is odd, then there is no solution to 2πk/n = π for any integer k = 0,1,2,...,n, so the only real root of x^{n}- a is u = uw^{0}.

if u < 0, then since n is odd, n = 2t +1, so u^{n}= (u^{2t})(u) = (u^{t})^{2}(u) < 0, since (u^{t})^{2}> 0.

hence u^{n}- a < 0, contradicting that u is a root of x^{n}- a. hence u > 0.

(i've left some of the details out but this is the general idea).