I know $\sqrt{2}$ is a solution to the equation $x^{x^{x^{x^{\vdots}}}}=2$. But why it doesn't blow out? I saw the trick to solve  $x^{x^{x^{x^{\vdots}}}}=2$, and solution is $\sqrt{2}$. I understand that $x^{x^{x}}$ is different from $(x^{x})^x$, but I still cannot see intuitively why the $x^{x^{x^{x^{\vdots}}}}$ doesn't blow out when $x$ is larger than 1.
I experimented a little, and found that all values that satisfy the equation are less than $\sqrt{2}$. What makes this range special?
$a=2:20$
$b=a^{\frac{1}{a}};b$
Thank you in advance for any help
 A: This question is analogous to asking the (equally good) question: why does the infinite series $\sum_{k=1}^\infty r^k$ blow up when $r>1$ but not when $0<r<1$? The answer, in both cases, is that the infinite expression is defined to be the limit of its finite truncations; and sometimes limits exist and sometimes they don't. In this case, the sequence
$$
x,\, x^x,\, x^{x^x},\, x^{x^{x^x}}, \dots
$$
happens to stabilize if $e^{-e}\le x\le e^{1/e}$ but not if $x>e^{1/e}$.
A: Maybe you're confusing $\sqrt{2}^{\left({\sqrt{2}}^{\sqrt{2}}\right)},\ $ with $\left({\sqrt{2}}^{\sqrt{2}}\right)^{\sqrt{2}}. $ The latter is done by calculating ${\sqrt{2}}^{\sqrt{2}}$ and then doing this number to the power of ${\sqrt{2}}.\ $
In fact, $\sqrt{2}^{\left({\sqrt{2}}^{\sqrt{2}}\right)},\ $ is ${\sqrt{2}}$ to the power of ${\sqrt{2}}^{\sqrt{2}},\ $ or to say it another way, you're doing ${\sqrt{2}}^{\sqrt{2}},\ $ and then doing ${\sqrt{2}}$ to the power of this number.
Expanding on this. Pretend ${\sqrt{2}} = 1.4.$
$1.4^{1.4} = 1.6.$
$1.4^{1.6} = 1.7.$
$1.4^{1.7} = 1.77.$
$1.4^{1.77} = 1.81.$
$1.4^{1.81} = 1.84.$
If the exponent is $< 2,\ $ then $1.4^{exponent} = new\_exponent$ will be less than $1.96$ which is less than $2$, so the next step in the iteration process is to do $1.4^{new\_ exponent}$, and this new exponent will again be be less than $1.96$ and so the result from the iteration process can never be greater than $1.96.$
Whereas,
$1.4^{1.4} = 1.6$
$1.6^{1.4}=1.93$
$1.93^{1.4}=2.5$
$2.5^{1.4}=3.6$
$1.93^{1.4}=6.0$
and this is growing without bound.
