Convergence of tetration sequence. This question arose from here. I am interested to find a nice proof about the convergence of $${^n}a=\underbrace{a^{a^{\ .^{\ .^{\ .^a}}}}}_{n\ \text{times}}.$$
I find with google a necessary and sufficient condition to have the convergence is $\frac{1}{e^e} \leq a \leq e^{1/e}$ but the part for $a\le1$ need some ugly work. 

Does anyone have an elegant/slick proof ?

 A: 
Does anyone have an elegant/slick proof ?

Like in the answer given to the link you quote by Kiril, the fastest proof is given by fixed point iteration. The function you iterate, is:
$$g_c(x)=c^x$$
The fixed points of $g_c(x)$ are given in terms of the Lambert $W$ function as:
$$x_0=\frac{W(-\ln(c))}{-\ln(c)}$$
The fixed point condition (for attraction hence convergence) is therefore:
$$|g'(x_0)|\le 1\Rightarrow$$
$$|-W(-\ln(c))|\le 1 (*)$$
Now consider the function:
$$m(x)=x\cdot\exp(x)$$
$$(*)\Rightarrow W(-\ln(c))\in [-1,1]\Rightarrow$$
$$m(W(-\ln(c)))\in m([-1,1])\Rightarrow$$
$$m(W(-\ln(c)))\in [-e^{-1},e]\Rightarrow$$
$$-\ln(c)\in[-e^{-1},e]\Rightarrow$$
$$c\in[e^{-e},e^{1/e}]$$
Fixed point iteration, therefore, takes care of all cases where $c\in(e^{-e},e^{1/e})$ (open interval). The two end points have to be checked separately, because the fixed point condition is inconclusive there, being possibly $\pm1$.
The upper end point ($e^{1/e}$) is easy using standard Calculus (Consider the sequence $a_n=g_{e^{1/e}}^{(n)}(1)$ and prove that it converges to $e$, etc.). The lower end point ($e^{-e}$), is a bit tricky, but can also be done using Calculus, and you are done.
A: Firstly if $$a \le -1$$ then we are guaranteed $$1 \le |a^a| \le 0$$  so if the value converges it is definitely bounded 
For $$-1 < a \le 0$$ we have that $$ |a^a| \rightarrow 1 $$ so again boundedness isn't an issue
For $$0 < a \le 1$$ the same reasoning holds to ensure boundedness
Now for convergence notice the following. If we assume the tower converges to a finite value, that is:
$$ a^{a^{a^{a^{\vdots}}}} = n$$
Then we can take logarithms on both sides and use the power to product rule to find
$$ a^{a^{a^{a^{\vdots}}}}\log(a) = \log(n)$$
In other words:
$$n \log(a) = \log(n)$$
Exponentiate both sides
$$ a^n = n$$
Yielding:
$$a = n^{\frac{1}{n}}$$
Now we can use calculus to find the maximum and minimal values that $n^{\frac{1}{n}}$ take on and this gives us the range of possible $a$ among the real numbers if we want $n$ to be among the real numbers.
