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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
$2$.