TITLE: The limit of $\lim\limits_{n\to\infty} \exp(-1+\exp(-2+\exp(-3+\cdots\exp(-n) \cdots)))$.
QUESTION [14 upvotes]: Does the following limit exist ?
$$\lim_{n\to\infty} \exp(-1+\exp(-2+\exp(-3+\cdots\exp(-n)\cdots)))$$

If yes, can it be expressed in a closed form ?
PARI shows the following numerical value :
n=-100;x=exp(n);while(n<-1,n=n+1;x=exp(x+n));print(x)

$0.4241685586940448516119410516$
Within this precision, $-200$ yields the same value.

REPLY [8 votes]: The limit exists because the sequence is monotonic and bounded, although I don't know yet what the limit is. Let's denote this sequence $E_n$.
The sequence is monotonic because $-n+\exp(-(n+1))>-n$, therefore $\exp(-n)+\exp(-(n+1)))>\exp(-n)$, therefore $\exp(-(n-1)+\exp(-n)+\exp(-(n+1))))>\exp(-(n-1)+\exp(-n))$, etc, until you get $E_{n+1}>E_n$.
The sequence is bounded because $-n+\exp(\text{negative number})<0$, therefore $E_n<\exp(0)=1$.
By monotone convergence theorem $E_n$ converges.