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 (xn) by x1=x3, xn+1=xxn. Observe that for x > 1, if x3 > 3 then the sequence is strictly increasing, and if x3 < 3 then the sequence is strictly decreasing.

Now to solve x(x3)=3, we are imposing x2=3, so the sequence becomes x3, 3, x3, 3, .... Since we must have x > 1 and the sequence is neither strictly increasing nor strictly decreasing, we must have x3=3. This also clearly works. So
x= 3 æ
Ö
3
 

.

Similarly, to solve x(x(x3))=3, we have x3=3, so the sequence becomes x3,x(x3), 3, x3, ... and so again we have
x= 3 æ
Ö
3
 

.

For the general equation
xxxxxx...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=±n1/n.