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**Bruno J.** We write $\displaystyle p(x)=\sum_{j=0}^n \alpha_jx^j$, with $\displaystyle \alpha_j \in F[x^n]$. We let $\displaystyle \epsilon$ be a primitive root of $\displaystyle x^{n-1}+x^{n-2}+\dots + 1 = 0$ in $\displaystyle \overline{F}$. Now, we let $\displaystyle q(x)=\prod_{1 \leq j<n} p(\epsilon^j x)$. A quick check shows that the coefficients of $\displaystyle q(x)$ lie in $\displaystyle F$. Now forget that the $\displaystyle \alpha_j$ are polynomials in $\displaystyle x$; the coefficients of $\displaystyle p(x)q(x)$, as a polynomial in $\displaystyle x$, are symmetric polynomials in $\displaystyle \epsilon^j \ (0 \leq j < n)$ with coefficients in $\displaystyle F[x^n]$. But since $\displaystyle \epsilon^n-1=0$, only the first and the last symmetric polynomial invariants do not vanish, and hence only powers of $\displaystyle x^n$ remain, i.e. $\displaystyle p(x)q(x) \in F[x^n]$...