$$\Large x^{x^{x^{x^{x^{.^{\,.^{\,.}}}}}}} = 2$$
What is $x$?
$$\Large x^{x^{x^{x^{x^{.^{\,.^{\,.}}}}}}} = 2$$
What is $x$?
So, we have $$x^2=2$$
In general, if $$x^{x^{x^{\cdots}}}=y, x^y=y$$
What you have there is called an infinite tetration. For your case, $x^2 = 2 \implies x = \sqrt2$.
In general, for $y = \Large x^{x^{x^{.^{\,.^{\,.}}}}}$, Euler showed that it is necessary that $e^{-e} \leq x \leq e^{\frac{1}{e}}$ for convergence to occur for real $x$.