TITLE: Limit of $\ln(1\cdot\ln(2\cdot\ln(3\cdot\ln(4\cdots))))$
QUESTION [9 upvotes]: I recently asked for the limit 
$$\lim_{n \to\infty} \ln(1+\ln(2+\ln(3+\ln(4+\cdots+\ln(n))\ldots)$$.
But what about the similar limit
$$\lim_{n \to\infty} \ln(1\cdot \ln(2\cdot \ln(3\cdot \ln(4\cdots \ln(n))\ldots)$$ ?
This limit also seems to exist
? n=10^5;x=log(n);while(n>1,x=log(x*(n-1));n=n-1);print(x)
0.3132776395465558314822583305

For n=100 the same value appears within this precision.
Is there a proof of convergence also for this limit ?
Can the limit be expressed by a closed form ?

REPLY [6 votes]: It's not a closed form, but there is a way to make this limit look even more interesting:
$$\lim_{n \rightarrow \infty} \ln(1\cdot \ln(2\cdot \ln(3\cdot \ln(4\cdots (\ln n) ))))=$$
$$
=\lim_{n \rightarrow \infty} \ln\ln\ln(3^2\cdot \ln^2(4\cdots(\ln n)))=
$$
$$
=\lim_{n \rightarrow \infty} \ln\ln\ln\ln^2(4^{3} \ln^3(5\cdots(\ln n) ))=
$$
$$
=\lim_{n \rightarrow \infty} \ln \ln \ln \ln^2 \dots \ln^{n-2} n^{n-1}=x
$$
Then, assuming that the limit $x$ exists, we can exponentiate $2$ times and get:
$$
\lim_{n \rightarrow \infty} \ln \ln^2 \ln^3 \dots \ln^{n-2} n^{n-1}=e^{e^x}
$$
Considering, that $n >> 1$, we can further simplify the expression:
$$
\lim_{n \rightarrow \infty} \ln \ln^2 \ln^3 \dots \ln^{n} n^{n}=e^{e^x}
$$
It's not immediately apparent that we are allowed to make this last simplification, but for some reason it appears to work numerically.