Rationalizing the denominator of $\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$ 
Simplify 
  $$\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}$$

I think this should be expressed without square roots at the denominator. I tried to multiply by conjugate.
 A: I used conjugates twice, got
$$ \sqrt 2 \cdot  \sqrt{2 + \sqrt 2} \cdot \sqrt{2 - \sqrt{2 + \sqrt 2}} $$
Perhaps these can be multiplied together to make something nice. I don't know. Note that no information about the source of the question is given. 
A: So, you multiply by the denominator over the denominator to get rid of the first square root (while still just technically multiplying only by 1):
$$
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}   \times \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\sqrt{2+\sqrt{2+\sqrt{2}}}} =
\frac{2\sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}
$$
Now you have an expression that can be multiplied by a conjugate, so that’s what we do:
$$
\frac{2\sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}\times\frac{2-\sqrt{2+\sqrt{2}}}{2-\sqrt{2+\sqrt{2}}}=\frac{\left(2\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}{(2-\sqrt{2})}
$$
This can also be multiplied by its conjugate:
$$
\frac{\left(2\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)}{(2-\sqrt{2})}=\frac{\left(2\sqrt{2+\sqrt{2+\sqrt{2}}}\right)\left(2-\sqrt{2+\sqrt{2}}\right)\left(2+\sqrt2\right)}{2}
$$
So, we’ve finally got an expression with a rationalized denominator.
I'd just leave the numerator unsimplified, as it's kind of hard to do and I know I'd definitely make a mistake (If your teacher/professor wants you to multiply out the numerator they're crazy;)).
A: $$
\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}}=\\\frac{2}{\sqrt{2+\sqrt{2+\sqrt{2}}}} \cdot \frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{\sqrt{2+\sqrt{2+\sqrt{2}}}}=\\\frac{2 \cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}}=\\
\frac{2\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2+\sqrt{2+\sqrt{2}}} \cdot \frac{2-\sqrt{2+\sqrt{2}}}{2-\sqrt{2+\sqrt{2}}}=\\\frac{2\cdot(2-\sqrt{2+\sqrt{2}})\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2}}=\\
\frac{2\cdot(2-\sqrt{2+\sqrt{2}})\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2-\sqrt{2}} \cdot \frac{2+\sqrt{2}}{2+\sqrt{2}}=\\\frac{2\cdot(2-\sqrt{2+\sqrt{2}})\cdot (2+\sqrt{2})\cdot \sqrt{2+\sqrt{2+\sqrt{2}}}}{2}=\\
(2+\sqrt{2}) \cdot (2-\sqrt{2+\sqrt{2}}) \cdot\sqrt{2+\sqrt{2+\sqrt{2}}}
$$
