This doesn't answer your question, but I thought I'd point out that if we assume $1 < a < b$, the solutions to $a^b = b^a$ can be written in the form $a = \left( \frac{s+1}{s} \right)^s$ and $b = \left( \frac{s+1}{s} \right)^{s+1}$ for some unique positive $s$. This is particularly useful when looking for rational solutions, since $s$ then has to be an integer (a homework exercise from BSM). Unfortunately, you're looking for a transcendental value (see D. Savitt's answer).

This doesn't answer your question, but I thought I'd point out that if we assume $1 < a < b$, the solutions to $a^b = b^a$ can be written in the form $a = \left( \frac{s+1}{s} \right)^s$ and $b = \left( \frac{s+1}{s} \right)^{s+1}$ for some unique positive $s$. This is particularly useful when looking for rational solutions, since $s$ then has to be an integer (a homework exercise from BSM). Unfortunately, you're looking for a transcendental value (see D. Savitt's answer).