Infinite sum of powers of the prime zeta function

denote by $\zeta_{p}(s)$ the Prime Zeta Function. Now, consider the infinite sum: $$D(s)=\sum_{n=0}^{\infty}\zeta_{p}(s)^{n}=\frac{1}{1-\zeta_{p}(s)}\;\;\;\left | \zeta_{p}(s)\right|<1$$ $D(s)$ could be thought of as a Dirichlet series in some half plane of convergence : $$D(s)=\sum_{n=1}^{\infty}\frac{a_{n}}{n^{s}}$$ Using the fundamental theorem of arithmetic, we can prove that: $$a_{n}\neq 0,\forall n\in \mathbb{Z}^{+}$$ Now, I'm having a hard time making sense of the coefficients $a_{n}$. In particular, they seem to bear some interesting arithmetical information with respect to the factorization of their corresponding indices. I have the feeling that I'm missing something basic !!

Any insights on the numbers $a_{n}$ are appreciated.

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I think that if $n$ can be written as a product of primes $n = p_1 \ldots p_r$ (here the primes factors are not necessarily distincts), then $a_n = r!$ (only $\zeta_p(s)^r$ contributes to the $n^{-s}$ term, and it contributes for $1$ for each permutation of the $p_i$). – Joel Cohen Nov 28 '12 at 20:22
@JoelCohen let me get this straight. you're saying that if $n$ is given by $n=\prod_{i}p_{i}^{b_{i}}$, then $a_{n}=\left(\sum_{i}b_{i}\right)!$, is that right !? – Mohammad Al Jamal Nov 28 '12 at 20:34
I think so, yes. My reasoning is in the comment above, but I might wrong... – Joel Cohen Nov 28 '12 at 20:39

Write $P(s)$ for the prime zeta function. By the multinomial theorem,
$$D(s)=\sum_{r=0}^\infty P(s)^r=1+\sum_{r=1}^\infty\sum_{e_{\large i}}\sum_{p_{\large i}~\mathrm{dstct}}{r\choose e_1,\cdots,e_r}\frac{1}{(p_1^{e_1}\cdots p_r^{e_r})^s}.$$
So the Dirichlet coefficients of $D(s)$ are given by multinomial coefficients in $p$-adic orders:
$$a_n={\displaystyle\sum_p v_p(n)\choose v_2(n),v_3(n),v_5(n),\cdots}.$$
(This expression makes sense because only finitely many $v_p(n)$ are nonzero for any natural $n$.)