When does $(x^x)^x=x^{(x^x)}$ in Real numbers? I have tried to solve this equation:$(x^x)^x=x^{(x^x)}$ in real numbers I  

got only $x=1,x=-1,x=2$ , are there others solutions ?

Note: $x$ is real number .

Thank you for your help .
Just a sketch for $x&gt;0$:

Let $y=x^x$ and rewrite $x^y=y^x$ by taking $log$'s as $\log (x)/x = \log(y)/y$.

For each $x \in (1, e)$ there is a unique $y(x)$ such that $f(x) = f(y(x))$.

As a function of $y(x)$, you can show that it is increasing. So we are looking for solutions of $y(x) = x^x$. Because $g(x) = y(x)-x^x$ is increasing, it can have only one root. So there is only one positive $x$ that satisfies the equation with $x\neq 1$.

EDIT: For integer $x &lt; 0$ non-integer, things get complicated because we will involve complex numbers and raising numbers to complex powers.
You have $(x^x)^x = x^{x^2}$. Assuming $x \not \in \{-1,1,1\}$,
$$x^{x^2} = x^{x^x}$$
happens if and only if the exponents agree, i.e. $x^2 = x^x$ which under our assumptions is equivalent to $x=2$.