Is the following statement true?
Let $x$ be algebraic over $\mathbb{Q}$ and let degree of $x$ over $\mathbb{Q}$ be n. Then, simple extension $\mathbb{Q}(x)$ will be equal to the set:
$$\{a_0 + a_1x + \dots + a_{n-1}x^{n-1} \mid a_i \in \mathbb{Q} \}$$
Now, my claim is that degree of $x^i$ (over $\mathbb{Q}$) for $1 \leq i \leq n-1$ is equal to $n$ if $i$ does not divide $n$ and $n/i$ otherwise. Furthermore, the degree of any combination $a_{i_1}x^{i_1} + \dots + a_{i_k}x^{i_k}$ would be equal to the least common multiple of all the degrees of single summands. I'm pretty convinced that it works for roots, but I'm not sure about complex numbers or sum of roots etc.
Neither claim is true (although the first claim, as corrected in Berci's comment, is true for roots of rational numbers).
For the first claim, nearly any non-root example falsifies it: take for example $x$ to be a root of $x^6+x+1$. The minimal polynomials of $x,x^2,\dots,x^{10}$ are $ \left\{x^6+x+1,x^6+2 x^3-x+1,x^6+3 x^4+3 x^2+x+1,x^6-2 x^3+4 x^2-x+1,x^6+5 x^5+10 x^4+10 x^3+5 x^2+x+1,x^6+6 x^5+15 x^4+20 x^3+15 x^2+5 x+1,x^6-7 x^3+14 x^2-6 x+1,x^6+8 x^4-2 x^3+12 x^2+7 x+1,x^6+3 x^4+18 x^3+12 x^2-8 x+1,x^6-5 x^5+10 x^4-8 x^3+25 x^2+9 x+1\right\}; $ in particular, all of those powers of $x$ have degree $6$ over $\Bbb Q$.
For the second claim, let $x=e^{2\pi i/n}$; then $x+x^2+\cdots+x^{n-1} = -1$ has degree $1$ over $\Bbb Q$.