Asymptotics of the solution of $x^x=n$ I need to find asymptotics of the solution of the equation 
$$
x^x=n
$$
while $n\to\infty$. The only thing I understand is this solution grows very slowly. I can't find $x$ explicitly, I think this is impossible. So what is the necessary trick?
 A: How about
$$
x =
\frac{\operatorname{ln} (n)}{\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)} + \frac{\operatorname{ln} (n) \operatorname{ln} \bigl(\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)\bigr)}{\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)^{2}} - \frac{\operatorname{ln} (n) \operatorname{ln} \bigl(\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)\bigr)}{\operatorname{ln} \bigl(\operatorname{ln} (n)\bigr)^{3}}  +\dots
$$
You can find computaiton of asymptotics for the Lambert
W function as Problem 4.2 in 
my paper Transseries for Beginners
A: As the other posters already pointed out, the Lambert W function is your friend here. If you take log on both sides then you get
$$
  x \log(x) = \log(n)
$$
If you assume that $x > 0$, then you can write $x = e^y$ (for $y = \log x$). Then it becomes
$$
e^y y = \log n
$$
which has $y = W(\log n)$ as a solution, where $W$ is the (principal branch of the) Lambert W function. Thus the solution to your original problem is $x = e^{W(\log n)}$, which grows to $\infty$ as $n \rightarrow \infty$ (although very, very slowly: $e(W(\log(10^6))) \approx 7$).
And you can even simplify this solution since (by definition)
$$
e^{W(\log n)} W(\log n) = \log n 
$$
and therefore
$$
x = e^{W(\log n)} = \frac{\log n}{W(\log n)}.
$$
