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.

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;)).
