The sum of infinite fours: $\sqrt{4^0+\sqrt{4^1+ \sqrt{4^2+ \dots}}}=?$ 
$\sqrt{4^0+\sqrt{4^1+\sqrt{4^2+\sqrt{4^3+\cdots}}}}=?$

I found this problem in a book. I tried to solve this but couldn't. Using calculator, I found the value close to $2$. But how can this problem be solved with proper procedure?

Take $f(x,n)=x+2^n$. We can see that;
$$\begin{aligned}
 f(x,n) &= \sqrt{2^{2n}+x\left(x+2^{n+1}\right)} \\
 &= \sqrt{2^{2n}+xf(x,n+1)} \\
  &= \sqrt{2^{2n}+x\sqrt{2^{2\left(n+1\right)}+x\sqrt{2^{2\left(n+2\right)}+x\sqrt{...}}}}\\ 
&=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\\
\end{aligned}$$
Taking $x=1,n=0$; we get;
$$2=\sqrt{4^{0}+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}$$
