If $x^{x^{x^{16}}}=16$ calculate the value of $x^{x^{x^{12}}}$ I have done the following but I'not satisfied.
if $x^a=a$ then by substitution follows that $x^a=x^{x^a}=x^{x^{x^a}}$ etc.
So $x^{x^{x^{16}}}=16$ is equivalent to $x^{16}=16$, and $x=2^{\frac{1}{4}}$  
Substituting $x$ to $x^{x^{x^{12}}}$ returns $2$.

It's obvious that if $x=\sqrt[16]{16}$, then $x^{16} = 16$, and $x^{x^{16}} =16$ and $x^{x^{x^{16}}} = 16$, so $x=\sqrt[16]{16}$ solves your equation. 
So then $x^{12} = {16^{0.75}} = 8$, and $x^{x^{12}} = x^8 = \sqrt{16} = 4$, and $x^{x^{x^{12}}} = x^4 = \sqrt[4]{16} = 2$. Hence $x^{x^{x^{12}}} = 2$

Ok hang on, I've got this beautiful approach!
Since $x^{x^{x^{16}}} = 16$, it follows that $x^{x^{x^{x^{x^{x^{16}}}}}} = 16$, and that $x^{x^{x^{x^{x^{x^{x^{x^{x^{16}}}}}}}}} = 16$, and so on , so forth, leading to the equation: $x^{x^{x^{x^{...}}}} = 16$. Now, if we look at $x$ exponentiated by both these equal exponents (I'm avoiding complex and negative values of $x$ then), then we get: $x^{x^{x^{x^{...}}}} = x^{16}$. From the previous two equations we get $x^{16}=16$, which give us $x=\sqrt[16]{16}$.  
