Proving $n^q$ is algebraic when $n\in \mathbb N$ and $q\in \mathbb Q$.
Definition: A number is called algebraic if it is the root of a polynomial: $P(x)=a_nx^n+...+a_1x+a_0$
My idea was to approach this problem by cases: $q\lt 0, q=0, q\gt 0$
If $q=0$, then we have $1$ which is certainly algebraic.
Otherwise,
Let $q=\frac ab$ where $a,b \in \mathbb Z$
Then, $n^{\frac ab}=(n^a)^{(\frac{1}{b})}$. We know that $n^a$ is algebraic because it satisfies $x-n^a$
Here is where I get stuck. How can this be applied to $\large^\frac{1}{b}$ and used to show $\gt 0$ and $\lt 0$ cases hold true? Any direction would be appreciated. Or, is this approach unfeasible?
Here's a way to derive a polynomial:
If $q=a/b$ where $a$ and $b$ are positive coprime integers, then set
$x = n^{a/b}$ which leads to $x^b = n^a$ and so $x^b-n^a = 0$
if $q=-a/b$, then set
$x=(1/n)^{a/b}$ which leads to
$x^b=(1/n)^a$ which leads to $n^ax^b - 1 = 0$.