Is $$ \sqrt{\sqrt{\sqrt{\sqrt{.....\sqrt x}}}} =1$$ where $x$ is a real number and $x > 0$?
Since $x$ after every under root , decreases exponentially I think it has to be $1$. But then $1^{2^{2^{2....^{2}}}} =1$ so I am confused.
I think the problem lies in the definition of the problem in the way, the expression is defined since the question can be reworded to $ \lim_{a \rightarrow \infty} x^{0.5^{a}} =1$.
It's not a standard notation. But your interpretation of it, in your final paragraph, is surely the only sensible one. And, given that interpretation, it is just as surely correct.
The $n$th iteration of the square root function is indeed $x\mapsto x^{2^{-n}}$, with inverse $y\mapsto y^{2^n}$. The limit $\lim_{n\to\infty}x^{2^{-n}}=1$, being a constant function, is not invertible. Similarly, the limit $$\lim_{n\to\infty}y^{2^{n}}=\begin{cases}\infty&y>1\\1&y=1\\0&y<1\end{cases}$$ is non-invertible too.
This should not confuse you. There is no reason to expect the limit of invertible functions to be invertible, and this is a prime example. (A simpler example is $x\mapsto ax$, with the limit taken as $a\to0$.)
Given $x_0 := x$ and $x_n>0$, update $x_{n+1} := \sqrt{x_n}$.
Case 1: If $x_0\geq 1$, then $1\leq x_{n+1} \leq x_n$.
Case 2: If $x_0<1$, then $x_n<x_{n+1}<1$.
In either case you have a monotone sequence that is bounded. Hence $(x_n)$ converges, say to $\ell$.
Then $\ell \leftarrow x_{n+1} = \sqrt{x_n} \to \sqrt{\ell}$ and thus $\ell = \sqrt{\ell}$, which implies $\ell=1$.
Another attempt:
For $x>0:$
$y_n:=x^{(1/2^n)} >0.$
$\log(y_n) =\dfrac{\log(x)}{2^n}.$
$\lim_{n \rightarrow \infty} \log(y_n) =0$.
$\lim_{n \rightarrow \infty} y_n = $
$\lim_{n \rightarrow \infty} \exp(\log(y_n)) =$
$\exp(\lim_{n \rightarrow \infty} (\log(y_n))=$
$\exp(0) =1$,
since $\exp$ is a continuous function.
Let $a_{n+1}:=\sqrt{a_n}$ with $a_0=\sqrt x.$ You can easily prove that $(a_n)_n$ is monotonous (depending on the value of $x$: if $0<x<1$ then $(a_n)_n$ is strictly increasing, if $x>1$ then $(a_n)_n$ is strictly decreasing, and $x=1$ is a trivial case since for $x=1$ we have $a_n=1, \forall n \in \Bbb N$) and bounded ($0<x<1 \Rightarrow a_n<1, \forall n\in\Bbb N$, $x>1 \Rightarrow a_n>1, \forall n \in \Bbb N)$. You can use induction to prove those. Then $(a_n)_n$ is convergent, and let $L:=\lim_{n \to \infty}{a_n}.$ Now we have $a_{n+1}=\sqrt{a_n}$ / $\lim_{n \to \infty} \Rightarrow$ $L = \sqrt{L} \iff L^2-L=0 \iff L=0$ or $L=1.$ Obviously, $L \ne 0$, and therefore $\lim_{n\to\infty}{a_n} = L = 1.$
