Closure of quadratic extensions under taking powers I'll give a proof of the following.
THEOREM: Let $x = p+\sqrt{q}$, with $p,q \in \mathbb{Q}$, and $m$ an integer. Then $x^m = a+b\sqrt{q}$ with $a,b \in \mathbb{Q}$.
PROOF
Using the Binomial Theorem
$${x^m} = C_0^m{p^m} + C_1^m{p^{m - 1}}{q^{\frac{1}{2}}} + C_2^m{p^{m - 2}}{q^{\frac{2}{2}}} +  \cdots  + C_{m - 2}^m{p^2}{q^{\frac{{m - 2}}{2}}} + C_{m - 1}^mp{q^{\frac{{m - 1}}{2}}} + C_m^m{q^{\frac{m}{2}}}$$
Let $m= 2·j$ then
$${x^{2j}} = C_0^{2j}{p^{2j}} + C_1^{2j}{p^{2j - 1}}{q^{\frac{1}{2}}} + C_2^{2j}{p^{2j - 2}}{q^{\frac{2}{2}}} +  \cdots  + C_{2j - 2}^{2j}{p^2}{q^{\frac{{2j - 2}}{2}}} + C_{2j - 1}^{2j}p{q^{\frac{{2j - 1}}{2}}} + C_{2j}^{2j}{q^{\frac{{2j}}{2}}}$$
Grouping produces
$${x^{2j}} = \sum\limits_{k = 0}^j {C_{2k}^{2j}{p^{2j - 2k}}{q^k}}  + \sum\limits_{k = 1}^j {C_{2k - 1}^{2j}{p^{2j - 2k + 1}}{q^{k - 1}}\sqrt q } $$
But since every binomial coefficient is integer, and every power of $p$ and $q$ is rational then one has
$${x^{2j}} = a+b\sqrt{q} \text{ ; and } a,b \in \mathbb{Q}$$ where $b$ and $a$ are the sums.
If $m = 2j+1$ then
$$x^{2j+1} =(a+b\sqrt{q}) (p+\sqrt{q}) = c+d\sqrt{q}$$
which is also in our set.
(I don't know if the ring-theory tag is appropiate.)
 A: More conceptually and more generally, suppose $\alpha$ is the root of a monic polynomial over a ring $R$
$$\alpha^n\ =\ r_{n-1}\ \alpha^{n-1} +\:\cdots\:+r_1\ \alpha + r_0$$
Multiplying the above by $\alpha$, and using the above as a rewrite rule to replace $\alpha^n$ by the lower powers on the RHS, we deduce that $\alpha^{n+1}$ may also be expressed as on the RHS. By induction so too can every power of $\alpha$, hence $R[\alpha] = R + \alpha\ R +\:\cdots\:+ \alpha^{n-1}\ R$, i.e. every polynomial in $\alpha$ with coefficients in $R$ is equal to one of degree $< n$.   
Equivalently, if monic $g(\alpha)= 0 $ for $g(x)\in R[x]$ of degree $n$, then, by the Division Algorithm, any polynomial $f(\alpha)$ is equal to an $h(\alpha)$ of degree $< n$, where $h(x)$ is the remainder of $f(x)$ mod $g(x)$
$$ f(x)\ =\ q(x)\ g(x) + h(x)\ \ \Rightarrow\ \ f(\alpha)\ =\ h(\alpha)\ \ by\ \ g(\alpha) = 0 $$
Recall that the high-school long Division (with Remainder) Algorithm works for any monic polynomial over any ring, so the the above normal-form degree-reduction algorithm works for any element $\alpha$ that is a root of monic polynomial. 
Such $\alpha$ generalize pure radicals $\sqrt[n]{r},\ r\in R$ and are known as algebraic integers over $R$, a fundamental concept in number theory and algebra. 
A: It seems correct, but it's much longer-winded than it needs to be. You could do this easily by induction and without any need for the binomial theorem:
Let $a,b,c,d \in \mathbb{Q}$.
If we can write
$$(a+b\sqrt{q})(c+d\sqrt{q}) = r+s\sqrt{q} \qquad (*)$$
for some $r,s \in \mathbb{Q}$, then we're done by induction.
Why? Because this will tell you that any finite product of elements of the form $a+b\sqrt{q}$ will take the same form, and a special case of this is a positive integer power of an element of said form, e.g. $(p+\sqrt{q})^m$.
So expanding the left-hand side of $(*)$ tells you that choosing
$$r = ac + bdq\ ,\ s = ad+bc$$
works since $q$ is rational.
