There are two definitions of

2^{3^{4}}

. Definition 1 gives

(2^{3})^{4}

which is

2^{12}

and definition 2 gives

2^{(3^{4})}

which is

2^{81}

Similarly the values of $(\sqrt{2}^{\sqrt{2}})^{\sqrt{2}}$ and $\sqrt{2}^{(\sqrt{2}^{\sqrt{2}})}$ are not equal. The first of these is $f (f (\sqrt{2}))$ where $f (x) = x^{\sqrt{2}}$ ; the second of these is $g (g (\sqrt{2}))$ where $g (x) = (\sqrt{2})^{x}$ .

To see what happen if you iterate the functions many times you should now experiment, using your calculator or computer, by iterating both $f$ and $g$ in each case starting with the value $\sqrt{2}$ .

Using these two definitions, we think of

$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{.^{.^{.}}}}}}}$

(where the powers of root 2 go on for ever) as the limit as $n$ to infinity of the sequence

$x_{1}, x_{2}, x_{3}, \dots x_{n}$

where, according to the first definition, $x_{n + 1} = f (x_{n})$ , or equivalently,

$x_{n + 1} = x_{n}^{\sqrt{2}}$

and, according to the second definition, $x_{n + 1} = g (x_{n})$ , or equivalently,

$x_{n + 1} = (\sqrt{2})^{x_{n}}$

In both cases, if the limit exists, you will find it by putting $x_{n + 1} = x_{n} = x$ .