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 $a>1$?
Consider the sum $$\sum_{i=0}^{\infty}\sum_{j=i+1}^\infty\sum_{k=j+1}^\infty\frac{1}{3^i3^j3^k} = \sum_{i=0}^{\infty}\left(\frac{1}{3^i}\sum_{j=i+1}^\infty\left(\frac{1}{3^j}\left(\sum_{k=j+1}^\infty\frac{1}{3^k}\right)\right)\right) = $$$$\frac{1}{2}\sum_{i=0}^{\infty}\left(\frac{1}{3^i}\sum_{j=i+1}^\infty\left(\frac{1}{3^j}\cdot 3^{-j}\right)\right) = \frac{1}{2}\sum_{i=0}^{\infty}\left(\frac{1}{3^i}\sum_{j=i+1}^\infty\left(9^{-j}\right)\right) = $$ $$\frac{1}{2}\sum_{i=0}^{\infty}\left(\frac{1}{3^i}\cdot3^{-2i-2}\cdot\frac{9}{8}\right)= \frac{1}{16}\sum_{i=0}^\infty 3^{-3i} = \frac{1}{16}\cdot\frac{27}{26} = \frac{27}{416}$$ This is summing over all $i<j<k$, which is one of the six possible cases when $i,j,k$ are distinct. Thus, the above number is one sixth of the answer.
To be clear, each of the intermediate steps were simply using the formula for infinite geometric series.
Now let's go for the general case. Suppose your general formula works for some $n$. Similarly, consider the sum
$$\sum_{k_1=0}^\infty \sum_{k_2=k_1+1}^\infty \sum_{k_3=k_2+1}^\infty... \sum_{k_n=k_{n-1}+1}^\infty \frac{1}{a^{k_1}a^{k_2}...a^{k_n}} = $$ (replacing $k_2$ with $k_2' = k_2-k_1-1$) $$\sum_{k_1=0}^\infty \sum_{k_2'=0}^\infty \sum_{k_3=k_2'+k_1+2}^\infty... \sum_{k_n=k_{n-1}}^\infty \frac{1}{a^{k_1}a^{k_2'+k_1+1}...a^{k_n}} = $$ (now, replacing $k_3$ with $k_3' = k_3-k_1-1$) $$\sum_{k_1=0}^\infty \sum_{k_2'=0}^\infty \sum_{k_3'=k_2'+1}^\infty\sum_{k_4 = k_3'+k_1+2}^\infty... \sum_{k_n=k_{n-1}}^\infty \frac{1}{a^{k_1}a^{k_2'+k_1+1}a^{k_3'+k_1+1}...a^{k_n}} = $$ (repeating this until we define $k_n'$) $$\sum_{k_1=0}^\infty\sum_{k_2'=0}^\infty\sum_{k_3'=k_2'+1}^\infty...\sum_{k_n'=k_{n-1}'+1}^\infty \frac{1}{a^{nk_1+n-1}a^{k_2'}...a^{k_n'}} = \sum_{k_1=0}^\infty \frac{1}{a^{nk_1+n-1}}c =$$$$ \frac{ca^{1-n}}{1-a^{-n}} = \frac{ca}{a^n-1}$$ Where $c$ is simply your formula, with $n-1$ being the number of sums, without the factorial term. Just as we had to multiply by $6=3!$ above, here, we multiply by $n!$ to come to your formula.
That takes care of the inductive step. The base case should be straightforward. My apologies if any of this was unclear.
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
$$n!\sum_{i_1=0}^\infty\sum_{i_2=i_1+1}^\infty\sum_{i_3=i_2+1}^\infty\cdots \sum_{i_n=i_{n-1}+1}^\infty \frac{1}{a^{i_1+\ldots+i_n}} = \frac{n!}{\prod_{i=1}^{n-1}a^i-1}\sum_{i=0}^\infty \frac{1}{a^{ni_1}} = \frac{n!a^n}{\prod_{i=1}^{n}a^i-1}$$