limit of the sequence $(x_{n+1})=a^{x_n}$ with $x_1=a$

Question

We are to find the limit of the sequence $(x_{n+1})=a^{x_n}$, where $x_1=a$ for $a>0$.

Attempt:

Case (1): $0<a<1$

We have $\frac{1}{a}=(1+ |r_1|)>1$ for some real $r$.

Hence, $\frac{1}{a^a}=(1+ |r_1|)^a=1+a|r_1|+\frac{a(a-1)}{2}|r_1|^2+...>1$

Evidently, $1+a|r_1|> \frac{1}{a^a}>1$ [$f(x) =1+ax-(1+x)^a \implies f'(x)<0]$. Repeating, $ 1+a^n|r_1|> \displaystyle\frac{1}{a^{a^{a^{{{.}^{{.}^{.}}}}}}}>1$. But, $\lim|a|^n =1$.

$\therefore$ by Sandwich Theorem we have the result $\lim x_{n+1}=1$

Case (2): $a=1$

Proof is evident.

Case (3): $a>1$

$1> (1/a) \implies 1>(1/a)>(1/a)^a>0$. Clearly, $1/x_n>1/x_{n+1}>0$.

But I want to show that $0$ is indeed an infimum or in some other way, that $(x_n)$ diverges to $\infty$ for $a>1$.

Also, consider checking the other cases.

Answer 1

If the limit $L$ does exist, it is given by the solution of $$L=a^L\implies L=-\frac{W(-\log (a))}{\log (a)}$$ where appears Lambert function.

However, in the real domain, $W(t)$ requires $t \geq -\frac 1e$ that is to say $a \leq e^{\frac{1}{e}}$