Let $a$ be a complex number that is algebraic over $\Bbb Q$ and let $r$ be a rational number. Prove that $a^r$ is algebraic over $\Bbb Q$.
Let $r = \frac{p}{q}$ where $q \neq 0$. Then $h(x) = x^q - a \in \Bbb Q(a)[x]$ is a minimal polynomial in $\Bbb Q(a)$. Then because $[F(a^r):F]=[F(a^r):F(a)][F(a):F] = \deg h(x) \cdot \deg f(x)$, where $f(x)$ is the minimal polynomial in $\Bbb Q[x]$, this shows that it is a finite extension $\Rightarrow$ algebraic.
The fact that $a$ is complex seems to be irrelevant based on this proof, did I miss something in my proof?
The use of the complex numbers is simply a convenient way of fixing an ambient field. And you probably want $h(x)=x^q-a^p$, so that $h(a^r)=0$.
Also, $h$ need not be the minimal polynomial of $a^r$ over $\mathbb{Q}(a^p)$, but since $h(a^r)=0$ we can at least say that $[\mathbb{Q}(a^r):\mathbb{Q}(a^p)]\leq \mathrm{deg}(h)$. This is enough to finish the proof using the tower rule.