TITLE: Proving the closed form for $\sum_{k_1=0}^{\infty}\cdots\sum_{k_n=0}^{\infty}\frac{1}{a^{k_1+\cdots+k_n}}$, where $k_1 \neq\cdots\neq k_n$ and $a>1$?
QUESTION [9 upvotes]: I recently encountered a problem that requires us to sum the series
$$
\sum_{i=0}^{\infty} \sum_{j=0}^{\infty} \sum_{k=0}^{\infty} \frac{1}{3^i 3^j 3^k}
$$
given the condition that $i \neq j \neq k$. Upon generalizing the problem, I get this:
$$
\sum_{k_1=0}^{\infty}\sum_{k_2=0}^{\infty}\cdots\sum_{k_n=0}^{\infty} \frac{1}{a^{k_1+\cdots+k_2}} = \frac{n! \times a^n}{\prod_{i=1}^{n} (a^i - 1)}
$$
for $a>1$ and $k_1 \neq \cdots \neq k_2$, i.e. all indices are distinct at all times. Now the closed form expression (on the right hand side) for the infinite series has been obtained purely by guessing. However, I've verified that the equation works, through a computer program. The only task now left to do is to prove the formula, which I'm unable to do.

REPLY [3 votes]: After writing this I noticed it looks close to the approach in florence answer, so I leave it as CW to complement that answer as the induction step here is a bit different.

By symmetry the product can be written 
$$n!\sum_{i_1=0}^\infty\sum_{i_2=i_1+1}^\infty\sum_{i_3=i_2+1}^\infty\cdots \sum_{i_n=i_{n-1}+1}^\infty \frac{1}{a^{i_1+\ldots+i_n}}$$
Performing the sums one-by-one from right to left the summand changes to
$$\frac{1}{a^{i_1+\ldots+i_n}}\to \frac{1}{a^{i_1+\ldots+i_{n-2}}a^{2i_{n-1}}} \left(\frac{1}{a-1}\right)\to \frac{1}{a^{i_1+\ldots+i_{n-3} + i_{n-2}}a^{3i_{n-2}}} \left(\frac{1}{a^2-1}\right)\left(\frac{1}{a-1}\right)$$
The pattern we see, the summand after performing $k$ sums is $\frac{1}{a^{i_1+\ldots + i_{n-k-1}}a^{(k+1)i_{n-k}}}\frac{1}{\prod_{i=1}^k a^i-1}$,  can be proven by induction: 
$$\sum_{i_{n-k} = i_{n-k}+1} \frac{1}{a^{i_1+\ldots + i_{n-k-1}}a^{(k+1)i_{n-k}}}\frac{1}{\prod_{i=1}^k a^i-1} = \frac{1}{a^{i_1+\ldots + i_{n-k-2}}a^{(k+2)i_{n-k-1}}}\frac{1}{\prod_{i=1}^{k+1} a^i-1}$$
which is the induction hypotesis for $k+1$. Taking $k=n-1$ we get
$$n!\sum_{i_1=0}^\infty\sum_{i_2=i_1+1}^\infty\sum_{i_3=i_2+1}^\infty\cdots \sum_{i_n=i_{n-1}+1}^\infty \frac{1}{a^{i_1+\ldots+i_n}} = \frac{n!}{\prod_{i=1}^{n-1}a^i-1}\sum_{i=0}^\infty \frac{1}{a^{ni_1}} = \frac{n!a^n}{\prod_{i=1}^{n}a^i-1}$$