Nested Root Simple Property

for me, the expression is equivalent to $x^{\frac{1}{2}} \times x^{\frac{1}{4}} \times x^{\frac{1}{8}} \times x^{\frac{1}{16}} ....$ which is equivalent to $x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...}$ and we know $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}...$ converges to $1$ , therefore the main expresion coverges to $x^{1}$ which is $x$

$\color{#D61F06}{y}=\sqrt{\color{#3D99F6}{x}\color{#D61F06}{\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}}}}}}}}}$

$\color{#D61F06}{y}=\sqrt{\color{#3D99F6}{x}\color{#D61F06}{y}}$

Squring both sides.

$\color{#D61F06}{y}^2=\color{#3D99F6}{x}\color{#D61F06}{y}$

$\color{#D61F06}{y}=\color{#3D99F6}{x}$

$\color{#3D99F6}{x}=\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}}}}}}}}$

$n = \sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\cdots}}}}}}}$

$n^2 = x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\cdots}}}}}}}$

$n^2 = xn$

$n^2 - xn = 0$

$n(n - x) = 0$

$\therefore n = 0, \text{ } n - x = 0$

$n = 0,\text{ } x$

When $n$ is $0$ :

$0 = \sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\cdots}}}}}}}$

$0^2 = x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\cdots}}}}}}}$

$x = 0$

$x \geq 0 \therefore x \neq 0$

Since $x \neq 0$ the only other solution is:

$x = \sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\cdots}}}}}}}$

Notice this example: $\sqrt{4\sqrt{16}}=\sqrt{4}\cdot\sqrt[4]{16}=4$ . So, as we add more roots, we divide by a higher power of 2. We can look at this as a geometric series: $\sum_{1}^{\infty}\frac{1}{2^{n}}$ . And we can find the convergence of a geometric series with: $\frac{a}{1-\left | r \right |}$ . If we do that, we get $\frac{0.5}{1-0.5}=1$ . So, $\lim_{x\rightarrow \infty}\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x\sqrt{x...}}}}}}}}}}}=x^{1}$ .