TITLE: $\alpha\in\mathbb{C}$ algebraic over $\mathbb{Q}$ iff $\alpha^n$ algebraic over $\mathbb{Q}$
QUESTION [5 upvotes]: Let $\alpha\in\mathbb{C}$ and $n\in\mathbb{N}$. Show, that $\alpha$ is algebraic over $\mathbb{Q}$, iff $\alpha^n$ is algebraic over $\mathbb{Q}$.

Hello, I want to proof this, but get stuck really quick...
I might need some help.
"$\Rightarrow$"
Suppose $\alpha$ is algebraic over $\mathbb{Q}$. Then there is $0\neq f\in\mathbb{Q}[X]$ with $f(\alpha)=0$. I have to show, that there is a $0\neq g\in\mathbb{Q}[X]$ with $g(\alpha^n)=0$
I tried severel things, but nothing of them should work. I want to construct $g$ by using $f$. 
First i thought about division with remainder. 
Then I tried to write $f(\alpha^n+\alpha-\alpha^n)$ and conclute something from there using the binomial theorem.
I also thought about using the degree of the expansion $[\mathbb{Q}(\alpha):\mathbb{Q}]=\deg(f)$, where we can suppose that $f$ is the minimal polynomial of $\alpha$.
Do you have a hint to get me started?
Thanks in advance.

REPLY [5 votes]: Here’s a much longer response identical in content to the comment of @JyrkiLahtonen:
You should know that $\alpha$ is algebraic over $\Bbb Q$ if and only if the field $\Bbb Q(\alpha)$ (the smallest field containing $\Bbb Q$ and $\alpha$) is finite-dimensional as a $\Bbb Q$-vector space.
Certainly $\Bbb Q(\alpha^n)$ is a subfield of $\Bbb Q(\alpha)$, so that if $\alpha$ is algebraic over $\Bbb Q$, then $\Bbb Q(\alpha^n)$ is too.