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.

Squaring both sides, you get$$3+\sqrt{13+4\sqrt3}=4+2\sqrt3,$$which is equivalent to $\sqrt{13+4\sqrt3}=1+2\sqrt3$. Squaring again both sides, you get $13+4\sqrt3=13+4\sqrt3$, and we're done.

Note that you can square at will, since we are dealing only with numbers greater than or equal to $0$.