TITLE: Evaluating sums of the form $\sum_{i_d=1}^{\infty}\ldots\sum_{i_2=1}^{\infty}\sum_{i_1=1}^{\infty}x^{i_1\cdot i_2\cdots i_d}$
QUESTION [5 upvotes]: I am wondering if there is a way to evaluate or get a more useful expression for a sum of the following form:
$$\sum_{i_d=1}^{\infty}\ldots\sum_{i_2=1}^{\infty}\sum_{i_1=1}^{\infty}x^{i_1\cdot i_2\cdots i_d},$$ where $|x|<1.$  For example, if $d=2$, then the sum is an example of a Lambert series and the exponents that appear are essentially given by (up to some index shuffling) this OEIS entry.  In this case it is not difficult to obtain $$\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}x^{ij}=\sum_{j=1}^{\infty}\frac{x^{j}}{1-x^{j}},$$ which I can then evaluate for any $|x|<1.$
This can be viewed as summing over a $d$-dimensional non-negative integer lattice, so I've looked at things like "lattice sums" (tried to post a link here but not enough reputation points to post more than 2 links) but can't seem to find anything helpful there.
Even something for the case $d=3$ would be helpful.

REPLY [2 votes]: Since you say you can evaluate the remaining infinite sum in the $d=2$ case, I gather that reducing the multiple sum to a single infinite sum would already be sufficient progress.
Then you want to count the number $\pi_d(n)$ of ways in which a positive integer $n$ can be written as an ordered product of positive integers. Find the prime factorisation, $n=\prod_ip_i^{k_i}$, and independently distribute $k_i$ balls over $d$ bins for each $i$, for a total of $\pi_d(n)=\prod_i\binom{k_i+d-1}{d-1}$ ways. Then your sum is
$$
\sum_{n=1}^\infty\pi_d(n)x^n\;,
$$
the ordinary generating function for $\pi_d(n)$. Note that $\pi_d(n)$ is a multiplicative function, which allows you to calculate it without factorising.