Another question from Titchmarsh's Riemann Zeta Function. I want someone to clarify me the next step in page 11.
They write for $\delta=\prod p^l$ ,$k=\prod p^{\lambda}$ the sum: $\sum_{\delta |k} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p|\delta}(l+1-lp^{-s})$
is
$$
\begin{align*}
k^{1-s}&\prod_{p|k}(\lambda+1-\lambda p^{-s})-\sum_{p|k} (\frac{k}{p})^{1-s}(\lambda-(\lambda-1)p^{-s})\prod_{p'|k, p'\neq p} (\lambda+1-\lambda (p')^{-s})\\
&+\sum_{pp'|k}(\frac{k}{pp'})^{1-s} (\lambda-(\lambda-1)p^{-s})(\lambda-(\lambda-1)(p')^{-s})\prod_{p'|k, p''\neq p,p'} (\lambda+1-\lambda(p'')^{-s})-...\\
&= k^{1-s}\prod_{p|k}((\lambda+1-\lambda p^{-s})-\frac{1}{p^{1-s}}(\lambda-(\lambda-1)p^{-s}))
\end{align*}
$$
If I understand correctly, the LHS of the above is from exclusion-inclusive principle, but I don't understand the RHS, I don't see how did they get the second term in the brackets of the product, i.e $\frac{1}{p^{1-s}}(\lambda-(\lambda-1)p^{-s})$.
Any hints?
Thanks.
 A: The notation used in the book is somewhat confusing as, although there is nothing to show it, $\lambda$ is a function of $p$, and $l$ is a function of $p$ and $\delta$.  To indicate this, I will write $l_\delta(p)$ and $\lambda(p)$ instead of $l$ and $\lambda$.
In the sum we start with, 
$$\Xi:=\sum_{\delta \mid k} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s}),$$ only values of $\delta$ for which $\mu(k/\delta)$ is nonzero contribute  to the sum.  Now, recall that $\mu(i)$ is nonzero only if $i$ is squarefree.  In this case, if $i$ has $j$ prime factors, $\mu(i)$ equals $(-1)^j$.  If $i$ is not squarefree, $\mu(i)$ is $0$.  Let the long sum starting $k^{1-s} \prod_{p\mid k}\ldots$ be called $\Sigma$, and call its first piece $\Sigma_0$, its second piece $\Sigma_1$, and so on, so that
$$\Sigma=\Sigma_0+\Sigma_1+\Sigma_2+\cdots.\qquad (*)$$  The expansion $(*)$ comes from setting $i:=k/\delta$ in $\Xi$ and looking successively at the cases where $j=0$, $1$, $2$, and so on.  For example, if $j$ is $0$, we must have $i=1$, so $\delta=k$ and
$$
\Sigma_0=\delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s})=
k^{1-s}\prod_{p\mid k}(\lambda(p)+1-\lambda(p)p^{-s}).
$$
This gives the first term of $\Sigma$.  The second term of $\Sigma$ comes from the case when $j=1$.  In this case, $i$ is prime, so $\delta=k/p$, for some $p$ dividing $k$.  Therefore, the contribution to $\Xi$ is
$$
\Sigma_1=\sum_{p\mid k, \delta=k/p} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s})$$
$$
=-\sum_{p\mid k} (\frac{k}{p})^{1-s} (\lambda(p)-(\lambda(p)-1)p^{-s})
\prod_{p'\mid k,\ p'\ne p}(\lambda(p')+1-\lambda(p')p'^{-s}).
$$
The third term of $\Sigma$ comes from the case where $j=2$.  In this case, $i=p p'$ for distinct primes $p$ and $p'$ dividing $k$, so $\delta=k/(p p')$.  The contribution to $\Xi$ is then
$$
\Sigma_2=\sum_{p, p'\mid k,\ p\ne p'} (\frac{k}{p p'
})^{1-s} 
(\lambda(p)-(\lambda(p)-1)p^{-s})
(\lambda(p')-(\lambda(p')-1)p'^{-s})$$
$$
\prod_{p''\mid k,\ p''\ne p,\ p''\ne p'}(\lambda(p'')+1-\lambda(p'')p''^{-s}).
$$
Continuing in this way, we can write down all of the expansion $(*)$.  This proves that $\Xi=\Sigma$.
To see that the product that we finally end up with, 
$$\Pi= k^{1-s}\prod_{p\mid k}((\lambda(p)+1-\lambda(p) p^{-s})-\frac{1}{p^{1-s}}(\lambda(p)-(\lambda(p)-1)p^{-s})),$$
is equal to $\Sigma$, we can write
$$\Pi=k^{1-s} \prod_{p\mid k} (A_p + B_p),\qquad (**)$$
where
$$A_p=\lambda(p)+1-\lambda(p) p^{-s},\ \ \ 
B_p=-\frac{1}{p^{1-s}}(\lambda(p)-(\lambda(p)-1)p^{-s}).$$
Then, if we expand $(**)$ and sum together all terms which contain exactly $j$ factors of the form $B_p$, we recover $\Sigma_j$.  For example, if we take $j=0$, we get $k^{1-s} \prod_{p\mid k} A_p$, which equals $\Sigma_0$.  If we take
$j=1$, we get
$$k^{1-s} \sum_{p\mid k} B_p \prod_{p'\mid k, \ p'\ne p} A_{p'},$$
which equals $\Sigma_1$, and so on.  This shows that $\Sigma=\Pi$.
Another way to derive $\Pi$ is to go back to the sum we started with,
$$
\Xi:=\sum_{\delta \mid k} \delta^{1-s}\mu(\frac{k}{\delta})\prod_{p\mid \delta}(l_\delta(p)+1-l_\delta(p)p^{-s}).
$$
If we rewrite this in terms of $\delta':=k/\delta$, and write $\delta'=\prod_{p\mid k} p^{l_{\delta'}(p)}$, we get
$$
\Xi=\sum_{\delta'\mid k} (\frac{k}{\delta'})^{1-s} \mu(\delta') 
\prod_{p\mid k} (\lambda(p)-l_{\delta'}(p)+1-(\lambda(p)-l_{\delta'}(p))p^{-s}).
$$
We can rewrite this as
$$
\Xi=k^{1-s} \prod_{p\mid k} (\lambda(p)+1-\lambda(p) p^{-s})
\sum_{\delta'\mid k} G(\delta'),\qquad (***)$$
where
$$G(\delta')=
\frac{1}{\delta'^{1-s}} \mu(\delta') 
\prod_{p\mid \delta'}\frac{\lambda(p)-l_{\delta'}(p)+1-(\lambda(p)-l_{\delta'}(p))p^{-s}}{\lambda(p)+1-\lambda(p) p^{-s}}.
$$
But now $G(\delta')$ is multiplicative, $G(1)=1$, and $G(\delta')$ vanishes unless $\delta'$ is squarefree.  Therefore
$$
\sum_{\delta'\mid k} G(\delta')=\prod_{p\mid k} (1 + G(p))
=\prod_{p\mid k} (1 - \frac{1}{p^{1-s}} 
\frac{\lambda(p)-(\lambda(p)-1)p^{-s}}{\lambda(p)+1-\lambda(p) p^{-s}}).
$$
Plugging this into $(***)$ gives $\Xi=\Pi$.
