Combinatorial Identity How does one prove the following identity in a combinatorial sense $$ \zeta(s) = \exp \left(\sum_{n=1}^{\infty} \frac{\Lambda(n)}{\log n} n^{-s} \right)$$
where $\zeta(s)$ is the Riemann zeta function and $\Lambda(n)$ is the Von Mangoldt function. By definition, $$\zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^{s}}$$
What does $\zeta(s)$ count? Can it be interpreted in terms of a probability? Also can be regard the right hand side as an exponential generating function?
 A: I have an idea for a proof, though I don't know if this qualifies to be a combinatorial proof at all. Still I'm giving this proof. Pardon me if it is not even close to what is desired.  
Since $$\Lambda(n)=\left\{\begin{array}{rl}
\log p & \mbox{if}\ n=p^k\\
0 & \mbox{else}
\end{array}
\right.$$
We can rewrite the RHS as the sum over all numbers of the form $p^k$ where $p$ are primes.
\begin{equation}
\begin{split}
\sum_{n\ge 1}\frac{\Lambda(n)}{\log n}n^{-s}=& \sum_{k\ge 1}\sum_{p}\frac{\Lambda(p^k)}{\log(p^k)}(p^k)^{-s}\\
\ =& \sum_{k\ge 1}\sum_{p}\frac{\log p}{k\log p}p^{-ks}\\
\ =& \sum_{k\ge 1}\sum_{p}\frac{1}{k}p^{-ks}\\
\end{split}
\end{equation}
Now, we can observe that, by Euler product \begin{equation}
\begin{split}
\zeta(s)=&\frac{1}{\displaystyle \prod_{p}\left(1-p^{-s}\right)}\\
\Rightarrow \log(\zeta(s))=& -\sum_{p}\log(1-p^{-s})\\
\ =&\sum_{p}\sum_{k\ge 1}\frac{p^{-ks}}{k}
\end{split}
\end{equation}
Hence $$\log(\zeta (s))=\sum\frac{\Lambda(n)}{\log n}n^{-s}\\
\Rightarrow \zeta(s)=\exp\left(\sum\frac{\Lambda(n)}{\log n}n^{-s}\right)  \Box $$
