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 $2^{3^4}$ either
as $(2^3)^4 = 2^{12}$ or as $ 2^{(3^4)} = 2^{81}$. In the same
way there are two interpretations of ${\sqrt 2}^{{\sqrt
2}^{\sqrt 2}}$ The first of these is $f(f(\sqrt 2))$ where
$f(x)=x^{\sqrt 2}$ which gives:
$({\sqrt 2}^{\sqrt 2})^{\sqrt 2} = {\sqrt 2}^{\sqrt 2 \times
\sqrt 2} = {\sqrt 2}^2 =2$
In the second case we get $g(g(\sqrt 2))$ where $g(x) = (\sqrt
2)^x$, and using a calculator to get an approximate value
gives:
${\sqrt 2}^{({\sqrt 2}^{\sqrt 2})} = {\sqrt 2} ^{1.63...} =
1.76 $ to $2$ decimal places.
So
${\sqrt 2}^{({\sqrt 2}^{\sqrt 2})}< ({\sqrt 2}^{\sqrt
2})^{\sqrt 2}.$
Now consider
${\sqrt 2}^{{\sqrt 2}^{{\sqrt 2}^{{\sqrt 2}^{{\sqrt 2}^
{{\sqrt 2}^{..}}}}}} $
where the powers of $\sqrt 2$ go on forever. We have seen that
we have two possibilities, namely
$X_1 = \lim x_n$ where $x_1 = \sqrt 2,\ x_{n+1}= x_n^{\sqrt
2}$ or
$X_2 = \lim x_n$ where $x_1 = \sqrt 2,\ x_{n+1} = (\sqrt
2)^{x_n}$.
N.B. Both iterations can be done on a calculator or computer:
$X_1 = \lim x_n$ is equivalent to iterating $f(x) = x^{\sqrt
2}$ and $X_2 = \lim x_n$ is equivalent to iterating $g(x) =
(\sqrt 2)^x$. If you do this experimentally, in each case
starting with $x_1=\sqrt 2$, 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
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 $X_2=2$.
Proof
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 $2$.