>>4
and in fact if f_n(x; a) = x^x^x^x ... ^x^a, then f_n(a^1/a; a) = a
(I don't know how this can be expressed using Knuth's up arrow notation
http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation but perhaps it is better expressed that way)
Knowing all this, your task now is to find the set of convergence of x^x^x^x^...
(hint: familiar constants will get in the way...)