Ho Chung gives a solution here:

For the first two equations, the answer is the cube root of 3, by observation. You can simply substitute this value in the equation and verify that it is a solution. We have to show that these are the only possible solutions.

What follows is really an argument by contradiction. If we assume that there exists a solution less than the cube root of 3 we reach an impossible situation and likewise for one greater than the cube root of 3.

Define a sequence $(x_{n})$ by $x_{1} = x^{3}$ , $x_{n + 1} = x^{x_{n}}$ . Observe that for $x > 1$ , if $x^{3} > 3$ then the sequence is strictly increasing, and if $x^{3} < 3$ then the sequence is strictly decreasing.

Now to solve $x^{(x^{3})} = 3$ , we are imposing $x_{2} = 3$ , so the sequence becomes $x^{3}$ , 3, $x^{3}, 3, . . .$ . Since we must have $x > 1$ and the sequence is neither strictly increasing nor strictly decreasing, we must have $x^{3} = 3$ . This also clearly works. So $x = \sqrt[3]{3}$ .

Similarly, to solve $x^{(x^{(x^{3})})} = 3$ , we have $x_{3} = 3$ , so the sequence becomes $x^{3}$ , $x^{(x^{3})}$ , 3, $x^{3}, . . .$ and so again we have $x = \sqrt[3]{3}$ .

For the general equation

$x^{x^{x^{x^{x^{x^{{...}^{n}}}}}}} = n$

where the sequence of powers is defined in the same way, and $n$ is a positive integer, we can use the same argument. The solution is the $n$ -th root of $n$ if $n$ is odd and when $n$ is even there are two solutions $x = \pm n^{1 / n}$ .