Specific theorem used to derrive $\frac{1}{\zeta(s)}=\sum_{n=0}^\infty\frac{\mu(n)}{n^s}$ In the book Prime Obsession by John Derbyshire he used the following reasoning to derive that identity
$$(a+b)(c+d) = ac+ad+bc+bd$$
$$(a+b)(c+d)(e+f)=ace+acf+ade+adf+bce+bcf+bde+bdf$$
The resulting sum of the product involves adding up all possible products where a term is grabbed from each parenthesis. He then applies that reasoning to the euler product:
$$\frac{1}{\zeta(s)}=\prod_{p}({1-p^{-s}})$$
When the product rule above is applied, only prime power terms remain in the sum, and each one is multiplied by either $1$ or $-1$ depending on the power. This gives:
$$\frac{1}{\zeta(s)}=\sum_{n=0}^\infty\frac{\mu(n)}{n^s}$$
The identity used looks like it can be generalized to the following formula:
$$\prod_{n=1}^k({a_n+b_n})=\sum{?}$$
Though I'm not sure what I should put into the sigma to represent the multiple combinations of the different terms that should end up in the sum (it might require nested sums?). Is there a specific theorem that generalizes the multiplication rule used to derive the identity? How exactly does such a theorem still work as we let $k->\infty$ (i.e under what conditions does it work for infinite products)? Is there a clearer way to see how the mobius function pops out of this?
 A: Let $p_n$ denote $n^{th}$ prime number.
Take $$P_k = \prod_{n=1}^k (1-a_n) $$
It is easy to see by induction that  $$P_k = 1 - \sum_{1 \le i_1 \le k}a_{i_1} +\sum_{1 \le i_1 < i_2 \le k}{a_{i_1}a_{i_2}} - \ldots+ (-1)^k a_1a_2\ldots a_k $$
where the general terms is  $$(-1)^r\sum_{1\le i_1 < i_2 < \cdots<i_r\le k}a_{i_1}a_{i_2}\cdots {a_{i_r}} $$
Now take $a_n = p_n^{-s}$,
then
$$(-1)^r\sum_{1\le i_1 < i_2 < \cdots<i_r\le k}a_{i_1}a_{i_2}\cdots {a_{i_r}} = (-1)^r \sum_{1\le i_1 < i_2 < \cdots<i_r\le k}p_{i_1}^{-s}p_{i_2}^{-s}\cdots {p_{i_r}^{-s}}$$
If you consider $m = p_{i_1}^{-s}p_{i_2}^{-s}\cdots {p_{i_r}^{-s}}$ then $\mu(m) = (-1)^r$, also see that there in the expansion of $P_k$ only squarefree terms appear and since $\mu(m) =0$ when $m$ is not squarefree, we can easily see that the coefficient of $m^{-s}$ in the expansion of
$$P_k = \prod_{n=1}^k (1-a_n) $$ is $\mu(m)$.
By taking $\lim_{k\to \infty}$ we get the desired result.
Update
We can write $$\prod_{n=1}^{k} (a_n+b_n) = \prod_{n=1}^{k} a_n \left(1+\frac{b_n}{a_n} \right) = (a_1 a_2 \ldots a_n) \prod_{n=1}^{k} (1 + c_n)  $$ where $c_n = b_n / a_n$ and we have already worked out $\prod_{n=1}^{k} (1 - (-c_n))$.
I am not sure if this identity has a particular name but a very similar expression appears in principle of inclusion and exclusion
A: The most broad-scope generalization of this is the Euler product formula for multiplicative functions.
Theorem (Euler product): Let $f(n)$ be a multiplicative arithmetical function such that $\sum_n|f(n)|$ converges. Then we have
$$
\prod_p\sum_{k\ge1}f(p^k)
$$
where $p$ runs over all positive prime numbers.
Proof. By fundamental theorem of arithmetic, we know that every positive integer $n$ can be written as a product of prime powers $n=\prod_pp^{r_p}$ where $r_p>0$ when $p|n$ and $r_p=0$ otherwise. By iterating through all $n\ge1$, we can just iterate through all possible combinations:
\begin{aligned}
\sum_{n\ge1}f(n)
&=\sum_{r_2,r_3,\cdots\ge0}f(2^{r_2}3^{r_3}\cdots)=\sum_{r_2\ge0}f(2^{r_2})\sum_{r_3\ge0}f(2^{r_3})\cdots \\
&=\prod_p\sum_{r_p\ge0}f(p^{r_p})=\prod_p\sum_{r\ge0}f(p^r)
\end{aligned}
Plugging $f(n)=n^{-s}$ in, we see that for $\Re(s)>1$ there is
$$
\zeta(s)=\sum_{n\ge1}{1\over n^s}=\prod_p\sum_{k\ge1}{1\over p^{ks}}=\prod_p\left(1-{1\over p^s}\right)^{-1}
$$
Moreover, by plugging in $f(n)=\mu(n)n^{-s}$, we observe that for $\Re(s)>1$ there is
$$
\sum_{n\ge1}{\mu(n)\over n^s}=\prod_p\left(1-{1\over p^s}\right)
$$
Combining these two results, we see that whenever $\Re(s)>1$ there is
$$
\sum_{n\ge1}{\mu(n)\over n^s}={1\over\zeta(s)}\tag{$*$}
$$
NOTE: The prime number theorem is equivalent to make ($*$) hold for $\Re(s)\ge1$, and the Riemann hypothesis is equivalent to make ($*$) hold for $\Re(s)>1/2$.
