Proving $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$. How would I show $\sqrt{3+\sqrt{13+4\sqrt{3}}} = 1+\sqrt{3}$?
I tried starting from the LHS, and rationalising and what-not but I can't get the result...
Also curious to how they got the LHS expression from considering the right.

$$ \sqrt{3+\sqrt{13+4\sqrt3}}= \sqrt{3+\sqrt{1+2\times2\sqrt{3}+(2\sqrt{3})^2}}$$ 
$$=\sqrt{3+\sqrt{(1+2\sqrt{3})^2}}$$
$$=\sqrt{3+1+2\sqrt{3}}$$
$$=\sqrt{(1+\sqrt{3})^2}$$
$$=1+\sqrt3$$
