Find the sum $\sum _{n=1}^{\infty}a_1a_2a_3...a_n $ where $a_{n+1}=\ln\frac{e^{a_n}-1}{a_n}$. Let $\{a_n\} $ be defined as follows: 
$a_1 \gt 0$ and $$a_{n+1}=\ln\frac{e^{a_n}-1}{a_n}$$  for $n\ge 1.$
Then the sum 
$$\sum_{n=1}^{\infty}a_1a_2...a_n $$
is _____________
My attempt:(By using hint )  
By Given condition 
$a_{n+1}-a_n=\ln\frac{e^{a_n}-1}{a_n}-lne^{a_n} $
$\Rightarrow a_{n+1}-a_n=\ln\frac{1-e^{-a_n}}{a_n}       
\quad    (1) $ 
Now $e^x \gt 1+x \space \forall x \in R \quad (2)$
$\Rightarrow  e^{-a_n} \gt 1-a_n $
From (1) 
$ a_{n+1}-a_n \lt \ln\frac{1-1+a_n}{a_n}=0  $
Also $a_{n+1} \gt \ln \frac{a_n}{a_n}=0 $ using (2) for $x=a_n$ 
So $\{a_n\} $ is a nonotone decreasing sequence bounded below. 
Let $lim_{n\rightarrow \infty}a_n=l $. Then by the given recurrence relation 
$l=\ln \frac{e^l-1}{l} $ 
$\Rightarrow e^l(l-1)+1=0 $ is satisfied by $l=0$ 
Hence$ \exists k\in N$s.t $ \forall n\ge k$ ,we have $a_n \lt 1$
Let $a_1a_2...a_{k-1}=p $
Then $\sum_{n=k}^{\infty}a_1a_2...a_n=p\{a_k+a_ka_{k+1}+...\} $
$\Rightarrow \sum_{n=k}^{\infty}a_1a_2...a_n \lt p\{a_k+a_k^2+...\}$ since $a_n$ is monotone decreasing .
$\Rightarrow \sum_{n=k}^{\infty}a_1a_2...a_n \lt p \frac{a_k}{1-a_k} $
So
$\sum_{n=1}^{\infty}a_1a_2...a_n \lt S+p\frac{a_k}{1-a_k} $ 
where $ S=\sum_{n=1}^{k-1}a_1a_2...a_n $
So the given sereis is convergent.
I would like to know if my workings are correct.I don't know how to find the actual sum.Also I have seen an answer here  (What is $\sum_{n=1}^{\infty}a_1 a_2...a_n$?) but unfortunately I can't understand it.Please help me in finding the sum.Thanks in advance.
 A: Note that for
$$
f(x)=e^x-1\tag1
$$
we have
$$
f(x)=x+xf\left(\log\left(\frac{e^x-1}x\right)\right)\tag2
$$
Let
$$
a_{k+1}=\log\left(\frac{e^{a_k}-1}{a_k}\right)\tag3
$$
Identity:

$$
f(a_1)
=\sum_{k=1}^n\prod_{j=1}^ka_j+f(a_{n+1})\prod_{j=1}^na_j\tag4
$$

Proof: Induction. It is trivially true for $n=0$. Suppose it is true for some $n$.
$$
\begin{align}
f(a_1)
&=\sum_{k=1}^n\prod_{j=1}^ka_j+f(a_{n+1})\prod_{j=1}^na_j\tag{5a}\\
&=\sum_{k=1}^n\prod_{j=1}^ka_j+(a_{n+1}+a_{n+1}f(a_{n+2}))\prod_{j=1}^na_j\tag{5b}\\
&=\sum_{k=1}^n\prod_{j=1}^ka_j+a_{n+1}\prod_{j=1}^na_j+a_{n+1}f(a_{n+2})\prod_{j=1}^na_j\tag{5c}\\
&=\sum_{k=1}^n\prod_{j=1}^ka_j+\prod_{j=1}^{n+1}a_j+f(a_{n+2})\prod_{j=1}^{n+1}a_j\tag{5d}\\
&=\sum_{k=1}^{n+1}\prod_{j=1}^ka_j+f(a_{n+2})\prod_{j=1}^{n+1}a_j\tag{5e}\\
\end{align}
$$
Explanation:
$\text{(5a)}$: inductive hypothesis
$\text{(5b)}$: apply $(2)$ and $(3)$
$\text{(5c)}$: distributive property
$\text{(5d)}$: collect $a_{n+1}$ into the products
$\text{(5e)}$: collect the product into the sum
Thus, it is true for $n+1$.
$\large\square$
The Mean Value Theorem says that there is a $\xi$ strictly between $0$ and $x$ so that
$$
e^\xi=\frac{e^x-1}x\tag6
$$
That is, $\log\left(\frac{e^x-1}x\right)$ is strictly between $0$ and $x$. Thus, $a_{k+1}$ is strictly between $0$ and $a_k$. Thus, $a=\lim\limits_{k\to\infty}a_k$ exists and satisfies $a=\log\left(\frac{e^a-1}a\right)$, but this can only happen if $a=0$.
Therefore, $a_k$ tends monotonically to $0$. Thus, we can take the limit of $(4)$ to get
$$
f(a_1)=\sum_{n=1}^\infty\prod_{k=1}^n a_k\tag7
$$
