Can someone prove that the inverse of $x^x$ is not an elementary function? I want to prove that the inverse of $f(x)=x^x$ is not an elementary function.
With elementary function I mean a function of one variable which is the composition of a finite number of arithmetic operations $+$, $–$, $\times$,$\div$, exponentials, logarithms, constants, and solutions of algebraic equations (a generalization of $n$'th roots).
I have no idea on how to do it, I would appreciate any help. Thanks!
 A: Equivalently, you want to prove that the Lambert W function $W$ is not elementary:
$f^{-1}(y) = \ln(y)/W(\ln(y))$.  This was asked and answered on MathOverflow.
A: We must be careful: A function has an inverse iff it is injective and surjective (bijective). You didn't specify the domain of your function $f$. Therefore we can assume the maximal real domain or the maximal complex domain. $f$ is not injective with these domains and doesn't have an inverse (inverse function) therefore.
If a function $f$ with $f(x)=x^x$ is bijective or not, depends on its domain and codomain. Each function can be decomposed into bijective subfunctions by decomposing its domain and/or codomain.
Our proof holds for all bijective subfunctions of $f$ with open domain and therefore for all partial inverses of $f$ with open domain.
$\ $
$$f(x)=x^x$$
From the definition of the inverse $f^{-1}$ of $f$, we get
$$f(f^{-1}(x))=x.$$
$y=f^{-1}(x)$:
$$f(y)=x$$
$$y^y=x$$
In the first part, we solve the equation.
In the second part, we show that the solution is not an elementary function.
In the last step, we have to show that the branches of Lambert W are not elementary functions.
1.) In the first step, we will solve the equation.
$$y^y=x$$
$$e^{\ln(y)\ y}=x$$
$$\text{Ln}(e^{\ln(y)\ y})=\text{Ln}(x)$$
$\text{Ln(z)}=\ln(z)+2k_1\pi i\ \ (k_1\in\mathbb{Z})$:
$$\ln(y)\ y=\ln(x)+2k_1\pi i\ \ \ \ \ \ (k_1\in\mathbb{Z})$$
$e^t=y$:
$$\ln(e^t)\ e^t=\ln(x)+2k_1\pi i$$
$$(t+2k_2\pi i)e^t=\ln(x)+2k_1\pi i\ \ \ \ \ \ (k_2\in\mathbb{Z})$$
$$(t+2k_2\pi i)e^{t+2k_2\pi i}=e^{2k_2\pi i}(\ln(x)+2k_1\pi i)$$
$e^{2k_2\pi i}=1$:
$$(t+2k_2\pi i)e^{t+2k_2\pi i}=\ln(x)+2k_1\pi i$$
Applying Lambert W:
$$t+2k_2\pi i=W_n(\ln(x)+2k_1\pi i)\ \ \ \ \ \ (n\in\mathbb{Z})$$
$$t=W_n(\ln(x)+2k_1\pi i)-2k_2\pi i$$
$t=\text{Ln}(y)$:
$$\text{Ln}(y)=W_n(\ln(x)+2k_1\pi i)-2k_2\pi i$$
$\text{Ln(y)}=\ln(y)+2k_3\pi i\ \ (k_3\in\mathbb{Z})$:
$$\ln(y)+2k_3\pi i=W_n(\ln(x)+2k_1\pi i)-2k_2\pi i$$
$$\ln(y)=W_n(\ln(x)+2k_1\pi i)+2k_4\pi i\ \ \ \ \ \ (k_4\in\mathbb{Z})$$
$$y=e^{W_n(\ln(x)+2k_1\pi i)+2k_4\pi i}$$
$$y=e^{W_n(\ln(x)+2k_1\pi i)}e^{2k_4\pi i}$$
$e^{2k_4\pi i}=1$:
$$y=e^{W_n(\ln(x)+2k_1\pi i)}$$
2.) In the second step, we will show that the solution is not an elementary function.
Assume our inverse $f^{-1}$ is an elementary function $E\ $ (1):
$$y=E(x)$$
$$e^{W_n(\ln(x)+2k_1\pi i)}=E(x)$$
$$\text{Ln}(e^{W_n(\ln(x)+2k_1\pi i)})=\text{Ln}(E(x))$$
$$W_n(\ln(x)+2k_1\pi i)=\ln(E(x))+2k_2\pi i\ \ \ \ (k_2\in\mathbb{Z})$$
$\ln(x)+2k_1\pi i\to t,x\to e^{-2k_1\pi i+t}$:
$$W_n(t)=\ln(E(e^{-2k_1\pi i+t}))+2k_2\pi i$$
The right-hand side of this equation is the function term of an elementary function (a composition of elementary functions), the left-hand side is the function term of the branches of Lambert W. If our assumption (1) is true, the branches of Lambert W are elementary functions.
3.) In the last step, we have to show that the branches of Lambert W are not elementary functions.
see e.g:
https://mathoverflow.net/questions/135911/how-to-prove-lamberts-w-function-is-not-elementary
Why is it that the Lambert W relation cannot be expressed in terms of elementary functions?
How can we show that A(z,e^z) and A(ln(z),z) have no elementary inverse?
