Simplify $\frac{1}{\sqrt{-1+\sqrt{2}}}$ In an exercise I got as solution $\frac{1}{\sqrt{-1+\sqrt{2}}}$, it now holds that $\frac{1}{\sqrt{-1+\sqrt{2}}} = \sqrt{1+\sqrt{2}}$. But I really don't see how you can manipulate the left hand side to become the right hand side (quite shameful as a mathematician, I must admit). 
Does anyone know how to do it?

You are given the fraction: $$\frac{1}{\sqrt{-1 + \sqrt{2}}} \tag1$$
Square it: $$\bigg(\frac{1}{\sqrt{-1 + \sqrt{2}}}\bigg)^2 = \frac{1^2}{\sqrt{-1 + \sqrt{2}}^2} = \frac{1}{-1 + \sqrt{2}} \tag2$$
Multiply by $1$: $$\frac{1}{-1 + \sqrt{2}}\times 1 = \frac{1}{-1 + \sqrt{2}}\times \frac{-1 - \sqrt{2}}{-1 - \sqrt{2}} = \frac{-1 - \sqrt{2}}{-1} \tag3$$
Now simplify (divide both terms in the numerator by $-1$): $$\frac{-1}{-1} - \frac{\sqrt{2}}{-1} = 1 - (-\sqrt{2}) = 1 + \sqrt{2} \tag4$$ And now finally, to cancel out the first square that we did in $(2)$, square root the $RHS$: $$\sqrt{1 + \sqrt{2}} \tag5$$
And there is you answer. You should end up with the following equation below: $$\frac{1}{\sqrt{-1 + \sqrt{2}}} = \sqrt{1 + \sqrt{2}} \tag{$\checkmark$}$$

EDIT:
The reason we make $1 = (-1 - \sqrt{2})/(-1 - \sqrt{2})$ is because $-1 - \sqrt{2}$ is the conjugate of $-1 + \sqrt{2}$. So pursuant to what @GeorgeN.Missailidis said, we did not need to square the fraction first, because all we could do was multiply it by $(\sqrt{-1 - \sqrt{2}})/(\sqrt{-1 - \sqrt{2}})$ since $\sqrt{-1 - \sqrt{2}}$ is the conjugate of $\sqrt{-1 + \sqrt{2}}$ anyway, but it is neater to cancel out the major roots.
