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Suppose $F$ is a field of cardinality $q$, and $n$ is an integer such that $n\mid q-1$. Now let $K$ be an extension of degree $n$ over $F$. Why does $x^n=\alpha$ have $n$ solutions in $K$ for any $\alpha\in F^\times$?

Since $K$ is finite of order $q^n$, its multiplicative group is cyclic, with generator $\gamma$. Writing $x=\gamma^y$ and $\alpha=\gamma^a$, I find that $$ x^n=\alpha\iff \gamma^{ny}=\gamma^a\iff ny\equiv a\pmod{q^n-1}. $$ The last congruence has solutions iff $(n,q^n-1)$ divides $a$, and in that case there are $(n,q^n-1)$ solutions. However, since $n\mid q-1$, and $q-1\mid q^n-1$, $n\mid q^n-1$ so $(n,q^n-1)=n$. So the only thing to prove is that $n\mid a$. I know that $\alpha^{q-1}=\gamma^{a(q-1)}=1$, and so $q^n-1\mid a(q-1)$. Hence $q^{n-1}+\cdots+q+1\mid a$, but I don't know how to continue. Why does $n\mid a$? Thanks.

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Try using the main result(s) in artofproblemsolving.com/Resources/Papers/LTE.pdf to prove in fact $\displaystyle n | \frac{q^n-1}{q-1}$. –  dinoboy Jan 23 '13 at 8:09
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Since $n\mid q-1$, you have $q\equiv 1\pmod{n}$. Then $$ q^{n-1}+\cdots+q+1\equiv \underbrace{1+\cdots+1+1}_{n\text{ times}}=n\equiv 0\pmod{n} $$ and thus $n$ divides $q^{n-1}+\cdots+q+1$, and so by transitivity of divisibility, $n\mid a$.

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