Prove that $ 3 =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{...}}}}$ Prove that $ 3 =\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{...}}}}$


According to the source, this can be done easily like
$$(x+1)^2 = 1+2x+x^2$$
$$(x+1)^2 = 1+x(x+2)$$

But how to continue and prove this?
Since $n^2 = 1+(n-1)(n+1)$, then $n = \sqrt{1+(n-1)(n+1)}$. 
Repeating the idea with $(n+1)^2 = 1+(n)(n+2)$, yields $n= \sqrt{1+(n-1)\sqrt{1+n(n+2)}}$. Further applications result in the infinite expression $n=\sqrt{1+(n-1)\sqrt{1+n\sqrt{1+(n+1)\sqrt{1+\cdots}}}}$.
Since $n^2 = 1+(n-1)(n+1)$, then $n = \sqrt{1+(n-1)(n+1)}$. 
Repeating the idea with $(n+1)^2 = 1+(n)(n+2)$, yields $n= \sqrt{1+(n-1)\sqrt{1+n(n+2)}}$. Further applications result in the infinite expression $n=\sqrt{1+(n-1)\sqrt{1+n\sqrt{1+(n+1)\sqrt{1+\cdots}}}}$.