Well done Paul Jefferys, you got close to a complete solution here. We have to consider two different values of these climbing powers depending on the order of operations which can be shown by putting in brackets. We can define 234 either as (23)4 = 212 or as 2(34) = 281. In the same way there are two interpretations of √2√2√2 The first of these is f(f(√2)) where f(x)=x√2 which gives:
(√2√2)√2 = √2√2 ×√2 = √22 = 2
In the second case we get g(g(√2)) where g(x) = (√2)x, and using a calculator to get an approximate value gives:
√2(√2√2) = √2 1.63... = 1.76  to 2 decimal places
So
√2(√2√2) < (√2√2)√2.
Now consider
√2√2√2√2√2√2..
where the powers of √2 go on forever. We have seen that we have two possibilities, namely
X1 = lim
xn

where x1 = √2, xn+1 = xn√2 or
X2 = lim
xn

where x1 = √2, xn+1 = (√2)xn. N.B. Both iterations can be done on a calculator or computer:
X1 = lim
xn

is equivalent to iterating f(x) = x√2 and
X2 = lim
xn

is equivalent to iterating g(x) = (√2)x. If you do this experimentally, in each case starting with x1=√2, you will find that the first iteration appears to converge to infinity and the second appears to converge to 2. We claim $X_1 = +\infty$. Proof \hfil\break We have $x_1=\sqrt{2}$ and $x_{n+1}=x_n^{\sqrt{2}}$; thus $$\log x_{n+1}=\sqrt{2}\,\log x_n, \quad \log x_1 = \log \sqrt{2}.$$ Thus $$\log x_{n+1} = \big(\sqrt{2}\big)^n\log \sqrt{2},$$ and as $\log x_n \to +\infty$ as $n\to \infty$, we see that $x_n\to +\infty$. We now claim that $X_2=2$. First we show that $x_n < 2$ for all $n$, and the proof is by induction. Clearly $x_1 < 2$. Now suppose that $x_n < 2$ and consider $x_{n+1}$. We have $$x_{n+1} = \big(\sqrt{2}\big)^{x_n} < \big(\sqrt{2}\big)^2 = 2$$ as required. Hence (by induction) $x_n < 2$ for all $n$. Next, we show by induction that $x_n < x_{n+1}$. It is clear that $$x_1 = \sqrt{2} < (\sqrt{2})^{\sqrt{2}} = x_2.$$ Now suppose that $x_{n-1} < x_n$. Then $${x_{n+1}\over x_n} = {\sqrt{2}^{x_n}\over \sqrt{2}^{x_{n-1}}} =\sqrt{2}^{x_n-x_{n-1}}.$$ Thus, as $x_n-x_{n-1}>0$, we have $x_{n+1}/x_n > 1$ and hence $x_{n+1}> x_n$. This shows that $x_n$ is increasing with $n$, and that $x_n <2$, and this is enough to see that $x_n$ converges to some number $X_2$, where $X_2\leq 2$. As $x_{n+1}=({\sqrt 2})^{x_n}$, if we let $n$ tend to infinity we see that $X_2$ is a solution of the equation $x=({\sqrt{2}})^x$. If we now plot the graphs of $y=x$ and $y=(\sqrt{2})^x$, we see that these two graphs meet at only two points, namely $(2,2)$ and $(4,4)$. Thus $X_2$ is either $2$ or $4$, and so it must be 2.