Is it possible that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$? Is it possible that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$ when $\mathbb{Q}(\alpha)$ is a $p$th degree Galois extension of $\mathbb{Q}$? 
($p$ is prime)
I got stuck with this problem while trying to construct polynomials whose Galois group is cyclic group of order $p$.
Edit: Okay, I got two nice answers for this question but to fulfill my original purpose(constructing polynomials with cyclic Galois group) I realized that I should ask for all primes $p$ if there is any such $\alpha$(depending on $p$) such that the above condition is satisfied. If the answer is no (i.e. there are  primes $p$ for which there is no $\alpha$ such that $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^{n})$ for all $n>1$) then I succeed up to certain stage.

If you mean for some $\alpha$ and $p$, then yes: if $\alpha=1+\sqrt{2}$, then 
$\mathbb{Q}(\alpha)$ is of degree 2, which is prime, and $\alpha^n$ is never a rational number, so $\mathbb{Q}(\alpha)=\mathbb{Q}(\alpha^n)$ for all $n>1$.
If you mean for all $\alpha$ such that the degree of $\mathbb{Q}(\alpha)$ is a prime number $p$, then no: if $\alpha=\sqrt{2}$, then $\mathbb{Q}(\alpha)$ is of degree 2, which is prime, but $\mathbb{Q}(\alpha^n)=\mathbb{Q}$ when $n$ is even.
