Theorem 1
Let $G \subset \mathbb{C}$ be an area and $\sum_{n=1}^{\infty}a_n e^{-\lambda _ns} $ and $\sum_{n=1}^{\infty}b_n e^{-\lambda _ns} $ two Dirichlet-series that converge on $G$ and represents the same funciton. Then $a_n = b_n$ for all $n$.
Theorem 2
Let $f: \mathbb{N} \to \mathbb{C}$ be a function and $F(s) = \sum_{n=1}^{\infty} f(n)n^{-s}$. Then $f$ is multiplicative if and only if $ \displaystyle F(s) = \prod_{p \in \mathbb{P}} \left( \sum_{l =0}^{\infty} f(p^l) p^{-ls} \right)$, where the product is absolut convergent.
We proofed Theorem 2 in the lecture and there is one step at the end (have a look at that first) that I don't understand. Here is an excerpt of the proof (proof from the right to the left).
For $N \in \mathbb{N}$ let $\{p_1, \ldots, p_r \} = \{ p \in \mathbb{P}; p \leq N \}$. It is
$\displaystyle F(s) = \sum_{n=1}^{N} f(n)n^{-s} = \sum_{p_1^{v_1} \cdots p_r^{v_r}\leq N} f(p_1^{v_1} \cdots p_r^{v_r}) \cdot (p_1^{v_1} \cdots p_r^{v_r})^{-s}$
and
$\displaystyle \prod_{p \leq N} \left( \sum_{l =0}^{\infty} f(p^l) p^{-ls} \right) = \sum_{p_1^{v_1} \cdots p_r^{v_r}\leq N} f(p_1^{v_1}) \cdots f(p_r^{v_r}) \cdot (p_1^{v_1} \cdots p_r^{v_r})^{-s} + \sum_{p_1^{v_1} \cdots p_r^{v_r} > N} f(p_1^{v_1}) \cdots f(p_r^{v_r}) \cdot (p_1^{v_1} \cdots p_r^{v_r})^{-s} $.
With the assumption we get
\begin{align}\displaystyle 0 &= \lim_{N \to \infty} \left(\sum_{n=1}^{N} f(n)n^{-s} - \prod_{p \leq N} \left( \sum_{l =0}^{\infty} f(p^l) p^{-ls} \right)\right) \\ &= \lim_{N \to \infty} \left ( \sum_{p_1^{v_1} \cdots p_r^{v_r}\leq N} \left[ f(p_1^{v_1} \cdots p_r^{v_r}) - f(p_1^{v_1}) \cdots f(p_r^{v_r}) \right]\cdot (p_1^{v_1} \cdots p_r^{v_r})^{-s} - \sum_{p_1^{v_1} \cdots p_r^{v_r} > N} f(p_1^{v_1}) \cdots f(p_r^{v_r}) \cdot (p_1^{v_1} \cdots p_r^{v_r})^{-s} \right) \end{align}
Since
$\displaystyle \sum_{p_1^{v_1} \cdots p_r^{v_r} > N} f(p_1^{v_1}) \cdots f(p_r^{v_r}) \cdot (p_1^{v_1} \cdots p_r^{v_r})^{-s}$
vanishes for $N \to \infty$, we have
\begin{equation}\displaystyle \sum_{r=1}^{\infty} \sum_{v_1,\ldots, v_r \in \mathbb{N}_0} \left[ f(p_1^{v_1} \cdots p_r^{v_r}) - f(p_1^{v_1}) \cdots f(p_r^{v_r}) \right]\cdot (p_1^{v_1} \cdots p_r^{v_r})^{-s} = 0 \end{equation}
Now here comes the step that isn't clear for me.
Using Theorem 1 we have, that all coefficients are 0.
My question is: Why can we apply Theorem 1? The left side of the equation resembles a Dirichlet-series but I don't see exactly why it is one and I need it to be one if I want to use Theorem 1...
Any help is much appreciated.
If the summation is absolutely convergent (something that is not immediately obvious to me), then you should be able to rearrange the summation into the form $$\begin{equation}\displaystyle \sum_{n=0}^{\infty} \left[ f(p_{1n}^{v_{1n}} \cdots p_{r_nn}^{v_{r_nn}}) - f(p_{1n}^{v_{1n}}) \cdots f(p_{r_nn}^{v_{r_nn}}) \right]\cdot (p_{1n}^{v_{1n}} \cdots p_{r_nn}^{v_{r_nn}})^{-s} = 0 \end{equation}$$
for some clever sequencing of the indices of the multiple summation. From there, just let $$a_n = f(p_{1n}^{v_{1n}} \cdots p_{r_nn}^{v_{r_nn}})$$ $$b_n = f(p_{1n}^{v_{1n}}) \cdots f(p_{r_nn}^{v_{r_nn}})$$ $$\lambda_n = \log (p_{1n}^{v_{1n}} \cdots p_{r_nn}^{v_{r_nn}}).$$