Understanding a calculation step involving Mobius function For $ n=p_{ 1 }^{k_{1}}p_{2}^{k_{2}}......p_{r}^{k_{r}} $ where $p_{i}$'s are prime numbers, define $P = \prod _ { p _ { i } | n } [ \mu ( 1 ) + \frac { \mu ( p _ { i } ) } { p _ { i } } + \cdots + \frac { \mu ( p _ { i } ^ { k _ { i } } ) } { p _ { i } ^ { k _ { i } } } ]$
According to the book Elementary Number Theory by David M. Burton, $P$ can be simplified as
$\frac { \mu ( 1 ) \mu ( p _ { 1 } ^ { a _ { 1 } } ) \mu ( p _ { 2 } ^ { a _ { 2 } } ) \cdots \mu ( p _ { r } ^ { a _ { r } } ) } { p _ { 1 } ^ { a _ { 1 } } p _ { 2 } ^ { a _ { 2 } } \cdots p _ { r } ^ { a _ { r } } }$ for $0 \leq a _ { i } \leq k _ { i }$.
Can anyone help me to understand how to achieve this form?
Definition of $\mu$ :
$\mu ( n ) = \left\{ \begin{array} { l l } { 1 } & { \text { if } n = 1 } \\ { 0 } & { \text { if } p ^ { 2 } | n \text { for some prime } p } \\ { ( - 1 ) ^ { r } } & { \text { if } n = p _ { 1 } p _ { 2 } \cdots p _ { r } , \text { where } p _ { i } \text { are distinct primes } } \end{array} \right.$
Please help....
 A: If you take a product of sums
$$\prod_{i=1}^n \sum_{j=0}^{k_i} u_{ij}$$
by repeated application of the distributive law, this becomes a sum of $\prod_{i=1}^n (k_i+1)$ terms. Each of these terms will be a product $\prod_{i=1}^n u_{ia_i}$ for some sequence of $n$ non-negative integers $a_i$ with $a_i \le k_i$ for all $i$. And for any such sequence $a_i$, the expanded expression will have the corresponding $\prod_{i=1}^n u_{ia_i}$ term. I.e.,
$$\prod_{i=1}^n \sum_{j=0}^{k_i} u_{ij} = \sum_{\{a_i\} \in \mathscr A} \prod_{i=1}^n u_{ia_i}, \quad \mathscr A = \{ \{a_i\} \subset \Bbb Z^{n}\mid \forall i, 0\le a_i \le k_i\}$$
This is how the second expression (with the missing summation symbol restored) was derived, with $u_{ij} = \dfrac{\mu(p_i^j)}{p_i^j}$.
There is an extraneous factor of $\mu(1)$ in the terms. Perhaps the author was including an $i=0$ factor of the original product which consisted of $\mu(1)$ only. But since this is just $1$, it doesn't matter.
As I noted in my comment, this is easily reducible by the properties of $\mu$. I don't know why the author has not already done so at this point. Presumably he makes that reduction later.
