To solve equations involving power towers (infinite tetration) we usually do something like this:
$$x^{x^{x^{x^{\dots}}}} =k$$
$$x^{(x^{x^{x^{\dots}}})} =k$$
$$x^k=k$$
$$x=\sqrt[k]k$$
But what if I do something like this:
$$x^{x^{(x^{x^{\dots}})}} =k$$
$$x^{x^k}=k$$
$$x^k\ln x=k$$
$$e^{k\ln x}\ln x=k$$
$$e^{k\ln x}k\ln x=k^2$$
$$x=e^{\frac {W(k^2)} k}$$
Where $W(z)$ is the Lambert W function.
This two expression must be equal so how can I prove this ? If they are not equal am I missing something whit the definition of infinite tetration ? And what about if I try to solve the equation doing this (and so on) ?:
$$x^{x^{x^{(x^{\dots})}}} =k$$
$$x^{x^{x^k}}=k$$
Actually you fixed $x$ and you're trying to find the limit so in the first case, if $k$ is the limit then $\sqrt[k]k=x$ and in the second case if $k$ is the limit then $x=e^{\frac{W(k\ln(k))}{k}}$ ( you made a mistake) but actually the two equations are equivalents : $$x=\sqrt[k]k\iff x=e^{\frac{\ln(k)}{k}}\iff x=e^{\frac{W(k\ln(k))}{k}}$$ using the fact that $W(k\ln(k))=\ln(k)$.
For higher order you will have the same expression, the limit is unique and can be proved rigorously under some conditions.