How to solve $x^{x^x}=(x^x)^x$? How can we solve the equation : 
$x^{x^x}=(x^x)^x$
with $x \in {\mathbb{R}+}^*$
Thanks for heping me :)

We can check when the exponents are equal.
It is $x^{x^x}=(x^x)^x\Leftrightarrow x^{(x^x)}=x^{x^2}\Leftrightarrow x^x=x^2$
Now x^x-x^2=0\Leftrightarrow $x^2(x^{x-2}-1)=0$.
So $x^2=0$ or $x^{x-2}=1$.
Since $x\neq 0$ we have $x^{x-2}=1$ left, which holds if $x-2=0$. So $x=2$
And $2^{2^2}=2^{2\cdot 2}$
Edit: And x=1 is an obvious solution...
