Abstract Algebra: x^p-a irreducible using automorphisms

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Homework Statement


Let F be a field with p\inN, a prime natural number. Show that either X^{p}-\alpha is irreducible in F[X] or \alpha has a pth root in F


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The Attempt at a Solution


I'm trying to do this without making reference to the field norm, so far I've come up with an incomplete proof that is missing some key components. First, if \alpha has a pth root in F then the polynomial (call it f) it is trivially reducible in F[X]. Assume then that there is no such thing in F. Assume also for a contradiction that f=gh, where g and h are not linear. There exists an extension field E in which f is a product of linear polynomials. So f=gh=(X-a_{1})(X-a_{2})...(X-a_{p}). Clearly, some of these linear factors will be factors of g and the rest factors of h. Suppose (X-a_{i}) is a factor of g and (X-a_{j}) not a factor of g. I want to say that there exists an F-automorphism (i.e. one that fixes F pointwise) that will take a_{i} to a_{j}. The coefficients in g will be unchanged by this automorphism, while the factors of g will not, suggesting that the field F is not a Unique Factorization Domain. This would be a contradiction and so f is irreducible. My issue is that I'm not sure whether such an automorphism necessarily exists and, if it does, what any of this has to do with p being prime, which I'm sure should be a key point in this proof. Any help would be much appreciated!
 
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Let ##K## be a splitting field of the polynomial ##f##.

Can you first show that every root of ##f## is distinct?

Second, can you show that ##K## contains all ##p##th root of unity?

Can you show then that if ##\zeta## is a ##p##th root of unity (which is necessarily in ##K##) and if ##\alpha## is a root of ##f##, then all roots of ##f## are given by

\{\alpha\zeta^n~\vert~0\leq n< p \}

Then if we can write ##f(X) = g(X)h(X)## with ##g## and ##h## polynomials with coefficients in ##F##, show that we can write

g(X) = \prod_{n\in S} (X- \alpha \zeta^n)~\text{and}~h(X) = \prod_{n\in S^c} (X-\alpha\zeta^n)

The constant term of ##g## is in ##F##, what is this constant term? Try to deduce that a root of ##f## lies in ##F##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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