$x^{x^{x^{\dots}}}=2$, what is the value of $x$? I came up with this little simple exercise, stating:

$x^{x^{x^{\dots}}}$ infinite times is equal to $2$, find $x$. 

As we're dealing with infinity, we can just separate the first $x$ and get $x^2=2 \implies x=\sqrt2$

But then I thought what if we change $2$ to $4$? Now we get $x^4=4 \implies x=\sqrt[4]4=\sqrt2$

Something went wrong, can someone explain?
One way to define the expression is $$y = x^{x^{x^{\cdots}}} $$

$$\Rightarrow y^x = x$$

A slightly different definition is to define the sequence $a_0 = x$ and $a_n = x^{a_{n-1}}$. The problem is then to find $$\lim_{n\rightarrow \infty} a_n$$

It can be shown that the interval of convergence for the power tower is $$[e^{-e}, e^{1/e}]$$
$x^{x^{x^x\cdots}}=2\implies x^2=2$