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.
 A: 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.
A: $\alpha=1+\sqrt 2$ (for $p=2$) should do the trick. By the binomial theorem, $(1+\sqrt 2)^n$ is always $a_n+b_n\sqrt2$ for some positive integer $b_n$.
A: If I understood Dinesh's intent correctly he wants an example for each prime $p$. Henning and Zev have covered the case $p=2$. 
Let first $p=3$.
Let $\zeta=e^{2\pi i/9}$. Then $L=\mathbf{Q}(\zeta)$ is the ninth cyclotomic field. It is well known that $Gal(L/\mathbf{Q})$ is cyclic of degree six. A generator $\sigma$ is determined by $\sigma(\zeta)=\zeta^2$. The subgroup generated by $\sigma^3$ has thus order two, and its fixed field $K=\mathbf{Q}(\zeta+\zeta^{-1})$ is a cyclic cubic Galois extension of $\mathbf{Q}$. I claim that $\alpha=\zeta+\zeta^{-1}=2\cos\left(2\pi/9\right)$ works. We can calculate that
$$
N^K_{\mathbf{Q}}(\alpha)=(\zeta+\zeta^{-1})(\zeta^2+\zeta^{-2})(\zeta^4+\zeta^{-4})=-1,
$$
so $\alpha$ is actually a unit of the ring of integers of $K$. But also note that $\alpha$
is a real number $>1$, so it cannot be a root of unity. Together these facts imply that no power of $\alpha$ will be rational. Namely, let $n>0$ be any integer.Then $\alpha^n$ is an algebraic integer of $K$, so if it were rational, it would have to be a rational integer $m$. On the other hand $N^K_{\mathbf{Q}}(\alpha^n)=(-1)^n$, so $m$ would have to be $\pm1$, and $\alpha$ would have to be a root of unity of order at most $2n$. A contradiction. Because $[K:\mathbf{Q}]=3$ is a prime, and $\alpha^n\in K\setminus\mathbf{Q}$, we have 
$K=\mathbf{Q}(\alpha^n)$ as desired.
Let then $p$ be any fixed prime $>3$. A way to generalize the above argument is to look at a bigger cyclotomic field. Let $\zeta=e^{2\pi i/p^2}$, and $L=\mathbf{Q}(\zeta)$. It is known that $L/\mathbf{Q}$ is a cyclic Galois extension of degree $\phi(p^2)=p(p-1)$. Let $H$ be the unique subgroup of this Galois group that is of order $p-1$. Then the fixed field $K$ of $H$ is a degree $p$ cyclic
extension of $\mathbf{Q}$. Furthermore, the restriction of the usual complex conjugation to
$L$ belongs to $H$, because it is of order two, and $2\mid p-1$. Therefore we can deduce that the numbers in the field $K$ are all real (the field $K$ is, in fact totally real). From Dirichlet's theorem of units we know that the group of units of the ring of integers of $K$ has a free abelian part of rank $p-1$. Let us pick a non-torsion element $u$ from that group. As $u$ is real, and not a root of unity, we can repeat the previous argument, and conclude that $K=\mathbf{Q}(u^n)$ for all positive integers $n$.
I would love to give you an explicit $u$ as in the case $p=3$, but I can't find a suitable
cyclotomic unit now that would belong to the field $K$. 
