How can one express $\sqrt{2+\sqrt{2}}$ without using the square root of a square root? I was trying to review some analysis, and came across problem 3 from page 78 of Walter Rudin's Principles of Mathematical Analysis.  As part of the problem, I wanted to try to write $\sqrt{2+\sqrt{2}}$ without using the square root of a square root.  In other words, I wanted to express the number in the form $q_1+q_2{q_3}^s$, where $q_1, q_2, q_3, s \in \mathbb{Q}$ (or perhaps something similar).  I'm aware that this is probably a duplicate of another question, but I wasn't able to find it (I wasn't sure what to search for)...  Many thanks in advance!
Edit (my work so far):
I tried expressing it as the solution to the quartic $x^4-4x^2+2=0$, but this seemed futile...
Edit 2 (the original problem):
The original problem from the text states:

If $s_1=\sqrt{2}$, and $$S_{n+1}=\sqrt{2+\sqrt{s_n}} (n=1,2,3,\ldots),$$
  prove that $\{s_n\}$ converges, and that $s_n<2$ for $n=1,2,3,\ldots$.

The problem is from the 3rd chapter of the book, which talks about sequences and series. Rudin provides numerous theorems on this topic, such as the comparison, ratio, and root tests for convergence.
 A: For completeness's sake here is the solution to the original problem: 
We first solve the equation
$$
\sqrt{2+\sqrt{x}}=x
$$
and the only real root $y$ is about 1.812. Now we have $\sqrt{2+\sqrt{2}}\approx 1.847$. If we differentiate $\frac{\sqrt{2+\sqrt{x}}}{x}$ with respect to $x$, the derivative is always negative. So when $s_{n}\ge y$, $s_{n}$'s value would decrease under iteration until it is less than $y$, and when $s_{n}\le y$ it would increase, since on the interval $[0,y]$ the quotient is greater than $0$. This force the limit to be $y$. 
Another way to see it is observing that 
$$
s_{n}=\sqrt{2+\sqrt{s_{n-1}}}, |s_{n+1}-s_{n}|=|\sqrt{2+\sqrt{s_{n}}}-\sqrt{2+\sqrt{s_{n-1}}}|=|\frac{\sqrt{s_{n}}-\sqrt{s_{n-1}}}{s_{n}+s_{n+1}}|
$$
therefore we have
$$
|\frac{s_{n+1}-s_{n}}{s_{n}-s_{n-1}}|=\frac{1}{|s_{n}+s_{n-1}||\sqrt{s_{n}}+\sqrt{s_{n-1}}|}\le \frac{1}{4}
$$
and the rest follows from contraction mapping theorem. 
A: Short answer: you can't. Call your element $\alpha$. You want $\alpha$ to live in a field extension of the form $\mathbb{Q}(\beta^{1/4})$ for some $\beta \in \mathbb{Q}$. It's clear that $L = \mathbb{Q}(\alpha)$ is a degree 4 field extension of $\mathbb{Q}$ (you've written down a minimal polynomial).
One can check easily that $L$ is Galois over $\mathbb{Q}$ (i.e. you can check that $\sqrt{2 - \sqrt{2}} \in L$). But $\mathbb{Q}(\beta^{1/4})$ (for $\beta$ squarefree) is not.
A: And then there is always trigonometry: $\sqrt{2+\sqrt{2}}=2\cos\left(\frac{\pi}{8}\right)$.
