The identity $\sum_n \ln(n) x^n = \sum_p ln(p)(\sum_k\frac{x^{p^k}}{1-x^{p^k}})$

As in the famous Euler product identity, the primes occur on only one side of the following:

$\sum_n \ln(n) x^n = \sum_p ln(p)(\sum_k\frac{x^{p^k}}{1-x^{p^k}})\ .$

My basic question: Does this identity appear in the literature?

If not, does the function $\sum_n \ln(n) x^n$? (It seems distantly related to polylogarithms.) Does it extend beyond the unit disk? (Computation suggests that behaves quit calmly up to the unit disk - I haven't detected visible evidence of a natural boundary.)

Is my(?) identity somehow equivalent to the Euler product identity?

Is here some obvious reason why it shouldn't be useful for studying the distribution of primes? (For a start it, does prove the infinitude of primes - if there were only finitely many primes, the coefficients on the left would have a bounded average.)

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For every two series $\{a _i\} _{i=1}^{\infty}$ and $\{b _i\} _{i=1}^{\infty}$, which satisfy $a _{n}=\sum _{k|n}b _k$ one has $$\sum _{n\geq 1}a _n x^n=\sum_ {n\geq 1} b _n \frac{x^n}{1-x^n}$$ this is known as "Lambert transformation". Therefore your identity is equivalent to the familiar statement $$\log(n)=\sum_{k|n}\Lambda(n)$$ with the von Mangoldt function. This identity is indeed equivalent to unique factorization and therefore to the Euler product identity. As to the literature, this identity is used in the beginning of the proof of the prime number thorem by Delange (1955), for example.
@Gjergji Yes. Or whether my expression might possibly contain any arithmetic information encoded in an analytic way not obviously available from the zeta function. If some comprensible transform turns $\sum_n \ln(n) x^n$ into the zeta function, I would take that as evidence for the answer "no." But I posted because I couldn't think of one myself. The summands $\sum_k\frac{x^{p^k}}{1-x^{p^k}}$ seem to behave badly at the boundary of the unit circle, but no so much the sum (seem = computationally, I have no proof), so I take that as a message that something is going on. –  David Feldman Jan 28 '11 at 22:47
I am not sure what you mean by comprehensible transform, but here is one: denote $l(x):=\sum_n \ln(n) x^n$. Then $\frac{1}{\Gamma(s)}\mathcal{M}\{l(e^{-t})\}(s)=-\zeta'(s)$, where $\mathcal{M}$ stands for the Mellin transform. This is a standard way to transform power series into Dirichlet series with the same coefficients. –  efq Jan 29 '11 at 2:52