Is ${\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=4$ correct? My teacher asked me this question. But I  think this is wrong. Anyone found this, before me? I do not know. Anyway, Is my solution correct?
${x^{x^{x^{x^x}}}}^{...}=2$
$x^2=2$
$x=\sqrt 2$
${\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=2$
Now,
Let, ${x^{x^{x^{x^x}}}}^{...}=4$
$x^4=4$
$x=\sqrt2$
${\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2^\sqrt2}}}}^{...}=4$
Contradiction!
 A: It's false: this number is defined as the limit of the sequence:
$$a_0=\sqrt2, \quad a_{n+1}={\sqrt{\rule{0pt}{1.8ex}2}}^{\,a_n}$$
It is easy to show by induction this sequence is increasing and bounded from above by $2$. Thus, by the monotone convergence theorem, it tends to a limit $\ell\le 2$, which satisfies the equation
$$\ell=\sqrt{\rule{0pt}{1.8ex}2}^{\,\ell}\iff \log\ell=\frac{\ell\log 2}2\iff\frac{\log \ell}\ell=\frac{\log 2}2$$
It is known this equation has two solutions: $\ell=2$ or $\ell=4$. As $\ell\le 2$, the limit is actually the former.
A: The problem is that the expression $x^{x^{\dots}}$ is ill-defined. What you want to do is take the limit of
$$x,x^x,x^{x^x},x^{x^{x^x}},\dots$$
For $x=\sqrt2$, this limit is indeed $2$. The reason you get $4$ is because the limit of the sequence
$$4,x^4,x^{x^4},x^{x^{x^4}},\dots$$
is $4$ when $x=\sqrt2$, and appears to look the same as the other sequence, however, they are not the same.
A: What's missing in your analysis is a pair of if's:
If there is a (positive real) number $x$ such that ${x^{x^{x^{x^x}}}}^{...}=2$, then $x=\sqrt2$.
and
If there is a (positive real) number $x$ such that ${x^{x^{x^{x^x}}}}^{...}=4$, then $x=\sqrt2$.
Both of these statements are true (you've proved that!).  So we can conclude that the function $f(x)={x^{x^{x^{x^x}}}}^{...}$ cannot have both $2$ and $4$ in its range.
Note, the only value of $x$ that is obviously in the domain of $f$ is $x=1$.  
To show that $\sqrt2^{\sqrt2^{\sqrt2^{\ldots}}}\not=4$, it suffices to show, by induction, that if $a_{n+1}=\sqrt2^{a_n}$ with $a_1=\sqrt2$, then $a_n\lt2$ for all $n$: The base case is $a_1=\sqrt2\lt2$, so if $a_n\lt2$ then $a_{n+1}=\sqrt2^{a_n}\lt\sqrt2^2=2$.
