Forgive the typos please, and sorry about the subject line. My intention was an infinite nesting of the exponents. The problem is to solve f(x) = x^{x^{x^{x^...ad infinitum...}}} = a. What makes it different from your x^x^3=3 is that there is no last exponent in the nesting. Otherwise the two have some similar features in their solutions. My problem has the solution x=a^(1/a) but with some caveats.
When a friend originally gave me the problem he had it with a=2. I figured it out when I realized that the exponent on the lowest x in the expression on the left side is the just the left side itself. Since the expression = 2, the exponent must be 2. Or x^2=2 and x=2^(1/2). Fooling around with the generalization, f(x)=a, with the solution x=a^(1/a), I tried a=4. Which would have the solution x=4^1/4 = 2^1/2. i.e. f(x)= 2 and f(x)= 4 have the same solution. f(2^(1/2))=2 and f(4^(1/4)) = 4, then 2=4 and I'm the pope, and everything else is true.
Actually, the problem only works for values of a between 1/e and e. (|ln(a)|<1.) And the infinite nested series only converges for values of x between 1/e^e and e^1/e. I always thought it was a fun problem. (Anytime you can prove that both of us are the pope, it's fun)

Paul