If $0<x$ is fixed, define the sequence $x_n$ recursively by $x_{n+1}=x^{x_n}$ and $x_0=x$. It is easy to show that if the limit $l$ of the sequence $x_n$ exists, then $(\frac{1}{e})^e\leq x \leq e^{\frac{1}{e}}$ and $x=l^{\frac{1}{l}}$. I found it much tougher to show that in that range of $x$, the limit does indeed exist. (This is an exercise in Knopp's book on infinite series).