Derivation of $\frac{1}{{p}_{1}\, {p}_{2}\, \cdots\, {p}_{m}} \prod\limits_{i = 1}^{m} \left({{p}_{i} - 1}\right)$ from reciprocal sum expansion How to show that 
\begin{equation*}
\begin{pmatrix}
1
- \sum\limits_{i = 1}^{m} \frac{1}{{p}_{i}} 
+ \sum\limits_{1 \le i < j}^{m} \frac{1}{{p}_{i}\, {p}_{j}}
- \sum\limits_{1 \le i < j < k}^{m} \frac{1}{{p}_{i}\, {p}_{j}\, {p}_{k}} + \cdots \\
+ \left({- 1}\right)^{m - 1} \sum\limits_{i = 1}^{m} \frac{{p}_{i}}{{p}_{1}\, {p}_{2}\, \cdots\, {p}_{m}}
+ \left({- 1}\right)^{m} \frac{1}{{p}_{1}\, {p}_{2}\, \cdots\, {p}_{m}}
\end{pmatrix} = \frac{1}{{p}_{1}\, {p}_{2}\, \cdots\, {p}_{m}} \prod\limits_{i = 1}^{m} \left({{p}_{i} - 1}\right)
\end{equation*}
I have not fully derived this and I am using the Principle of Inclusion-Exclusion arguments.  If this is incorrect please advise.
I suspect that there is a simple direct proof of this statement.
All the pi are distinct primes.
 A: First understand why the following is true:

$$\prod_{i = 1}^m (a_i + b_i) = \sum_S a_Sb_{S^c}$$
where the summation is over all subsets $S$ of $\{1,\dots,m\}$ and $x_T = \prod_{i \in T} x_i$. (For example, $a_{\{i,j\}} = a_ia_j$ and $a_{\{i,j,k\}} = a_ia_ja_k$ and $a_{\varnothing} = 1$.) $S^c = \{1,\dots,m\} \setminus S$ is the complement of $S$.

The reason for this is because for each factor of the product $(a_1 + b_1)\cdots(a_m + b_m)$ you select either the $a_i$ or the $b_i$. The set $S$ consists of the indices $i$ for which you selected $a_i$. You sum over all possible choices.
Note that in your case, with $a_i = p_i$ and $b_i = -1$, you have $b_{S^c} = (-1)^{|S^c|} = (-1)^{m - |S|}$. Also $p_1\dots p_m = p_{\{1,\dots,m\}}$, so what you have is
\begin{align}
\frac{1}{p_1\cdots p_m} \prod_{i=1}^m (p_i - 1) &= \sum_S \frac{p_S}{p_{\{1,\dots,m\}}} (-1)^{m - |S|} \\
&= \sum_S \frac{1}{p_{S^c}} (-1)^{m - |S|} \\
&= \sum_T \frac{1}{p_T} (-1)^{|T|}
\end{align}
making the change of variables $T = S^c$ at the end.
This is exactly what you have except you parameterize the sets $T$ of size $k$ as $T = \{i_1,i_2,\dots,i_k\}$ with $1 \le i_1 < i_2 < \cdots < i_k \le m$.
