I am stuck in the following puzzle and couldn't find a way to approach this.

$\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}}$

• Write $\sqrt{180} = a\sqrt{5}$, see that $14+\sqrt{180}$ is a square, continue in the same manner for the result in the next outer square root until you reach the end. – Daniel Fischer Oct 21 '14 at 10:11
• It evaluates to $\sqrt{5} + 1$. – James Harrison Oct 21 '14 at 10:12
$$\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{(\sqrt{5}+3)^2}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{5} + 3}} = \sqrt{5 + \sqrt{5} + \sqrt{6+2\sqrt{5}}} = \sqrt{5 + \sqrt{5} + \sqrt{(\sqrt{5}+1)^2}} = \sqrt{5 + \sqrt{5} + \sqrt{5}+1}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1$$
Hint: \eqalign{ & \sqrt{14 + \sqrt {180}} = 3 + \sqrt 5 \cr & \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr & \sqrt {5 + \sqrt 5 + \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr}