This is a generalization of my answer to $\sum_{1\leq l \lt m\lt n} \dfrac{1}{5^l3^m2^n}$
I showed there that, if $0 < a, b ,c < 1$,
$\begin{array}\\ s(a, b, c) &=\sum_{1\leq l \lt m\lt n} a^lb^mc^n\\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{n-1}\sum_{l=1}^{m-1}a^lb^mc^n\\ &=\dfrac{ab^2c^3}{(1-c)(1-bc)(1-abc)}\\ \end{array} $
Obviously
$\begin{array}\\ s(c) &=\sum_{1\leq n} c^n\\ &=\sum_{n=1}^{\infty}c^n\\ &=\dfrac{c}{1-c}\\ \end{array} $
and, with a little more work,
$\begin{array}\\ s(b, c) &=\sum_{1\leq m\lt n} b^mc^n\\ &=\sum_{n=1}^{\infty}\sum_{m=1}^{n-1}b^mc^n\\ &=\dfrac{bc^2}{(1-c)(1-bc)}\\ \end{array} $
This led to the following conjecture:
If $A_m = (a_1, ..., a_m) =(a_i)_{i=1}^m $ and $s(A_m) =\sum_{i_1=1}^{\infty} \sum_{i_2=1}^{i_1-1} \sum_{i_3=1}^{i_2-1} ... \sum_{i_m=1}^{i_{m-1}-1} \prod_{k=1}^m a_k^{i_k} $ then $s(A_m) =\dfrac{\prod_{k=1}^m a_k^{m+1-k}}{\prod_{k=1}^m (1-\prod_{j=1}^ka_j)} $.
I have a proof, but it is really cumbersome (as is just the statement of the problem and the result).
My hope is that someone can come up with a better statement of the sum and proof of the result.
There is a slightly different approach to evaluate $s(A_m)$ by using a relabelling of the indices. I illustrate it for the case $s(b,c)$, but it relatively easy to generalise. $$ s(b,c) = \sum_{1 \leq m<n} b^m c^n $$ If we introduce $\delta = n - m \geq 1$, we can rewrite the sum as $$ s(b,c) = \sum_{1\leq m} \sum_{1 \leq \delta} b^m c^{m+\delta} = \sum_{1 \leq m} (b c )^m \sum_{1\leq \delta} c^\delta = \frac{b c}{1 - b c} \frac{c}{1-c} $$ where I made the summation run over $\delta$. This of course gives the same final result that you already found.
You can use this approach to introduce differences between the summation indices $i_k$ and the final results becomes product of independent summations over unbounded positive indices and result in the expected product.
The only thing that remains to do, is to use induction to prove the final result, but I am sure you can do this yourself.