asymptotic of $x^{x^x} = n$ How find the asymptotic behavior for $x(n)$ 
if $x^{x^x} = n$?
I supposed that $x = O(\log\log{n})$
and took logarithm two times.
So I get
$x = O(\frac{\log\log{n}}{\log\log\log{n}})$
Is it right?
How to improve this estimate? 
 A: Let's write
$$
\begin{align}
&\log\log n = L_2(n), \\
&\log\log\log n = L_3(n), \\
&\log\log\log\log n = L_4(n).
\end{align}
$$
Then from $x^{x^x} = n$ we get
$$
x\log x + L_2(x) = L_2(n), \tag{1}
$$
and so, since $x \to \infty$ as $n \to \infty$, we have
$$
x\log x \sim L_2(n) \tag{2}
$$
as $n \to \infty$.  Taking logs of this yields
$$
\log x + L_2(x) \sim L_3(n),
$$
from which it follows that
$$
\log x \sim L_3(n). \tag{3}
$$
By dividing $(2)$ by $(3)$ we then find that
$$
x \sim \frac{L_2(n)}{L_3(n)}, \tag{4}
$$
agreeing with your estimate.  To get the next term we iterate.
Taking logs of $(1)$ and rearranging yields
$$
\log x = L_3(n) - L_2(x) - \log\left(1 + \frac{L_2(x)}{x\log x}\right). \tag{5}
$$
From $(3)$ we know that
$$
L_2(x) = L_4(n) + o(L_4(n)),
$$
so we can rewrite $(5)$ as
$$
\log x = L_3(n) - L_4(n) + o(L_4(n)).
$$
Substituting these into $(1)$ yields
$$
x\Bigl(L_3(n) - L_4(n) + o(L_4(n))\Bigr) = L_2(n) - L_4(n) + o(L_4(n)). \tag{6}
$$
Now
$$
\Bigl(L_3(n) - L_4(n) + o(L_4(n))\Bigr)^{-1} = L_3(n)^{-1} + \frac{L_4(n)}{L_3(n)^2} + o\left(\frac{L_4(n)}{L_3(n)^2}\right),
$$
so we can conclude from $(6)$ that
$$
x = \frac{L_2(n)}{L_3(n)} + \frac{L_2(n) L_4(n)}{L_3(n)^2} + o\left(\frac{L_2(n) L_4(n)}{L_3(n)^2}\right)
$$
as $n \to \infty$.
