Is $2\prod_{i=1}^{n}{p_i} - \prod_{i=1}^{n}{(p_i + 1)} - \prod_{i=1}^{n}{(p_i - 1)}$ even and negative for $n > 1$? Is $$2\prod_{i=1}^{n}{p_i} - \prod_{i=1}^{n}{\left(p_i + 1\right)} - \prod_{i=1}^{n}{\left(p_i - 1\right)}$$ even and negative for $n > 1$, where $p_i > 1 \hspace{0.07in} \forall i \in \left[1,n\right]$?
Here are a few quick WolframAlpha computational verifications:
$$2pq - (p+1)(q+1) - (p-1)(q-1) = -2$$
$$2pqr - (p+1)(q+1)(r+1) - (p-1)(q-1)(r-1) = -2(p+q+r)$$
$$2pqrs - (p+1)(q+1)(r+1)(s+1) - (p-1)(q-1)(r-1)(s-1) = -2\left(p(q+r+s)+q(r+s)+rs+1\right)$$
How do we prove the general case?
 A: To prove that it is even, it is sufficient to prove that
$$\prod_{i=1}^{n}(p_i+1)+\prod_{i=1}^{n}(p_i-1)\tag1$$
is even.
If $p_i+1\equiv 0\pmod 2$ for some $i$, then $p_i-1\equiv 0\pmod 2$, so $(1)= \text{even $+$ even $=$ even}$.
If $p_i+1\equiv 1\pmod 2$ for every $i$, then $p_i-1\equiv 1\pmod 2$ for every $i$, so $(1)=\text{odd $+$ odd $=$ even}$.
To prove that it is negative, use induction.
$$2p_1p_2 - (p_1 + 1)(p_2+1) -(p_1 - 1)(p_2-1)=-2\lt 0$$
Supposing that 
$$2\prod_{i=1}^{n}{p_i} - \prod_{i=1}^{n}{\left(p_i + 1\right)} - \prod_{i=1}^{n}{\left(p_i - 1\right)}\lt 0$$
gives
$$\begin{align}&2\prod_{i=1}^{n+1}{p_i} - \prod_{i=1}^{n+1}{\left(p_i + 1\right)} - \prod_{i=1}^{n+1}{\left(p_i - 1\right)}\\&=p_{n+1}\left(2\prod_{i=1}^{n}{p_i} - \prod_{i=1}^{n}{\left(p_i + 1\right)} - \prod_{i=1}^{n}{\left(p_i - 1\right)}\right)-\left(\prod_{i=1}^{n}{\left(p_i + 1\right)}-\prod_{i=1}^{n}{\left(p_i - 1\right)}\right)\\&\lt 0\end{align}$$
A: Yes, that expression is even and negative for $n > 1$.
It is even because $2\prod_{i=1}^{n}{p_i}$ is even, and $p_i+1\equiv p_i-1 \bmod 2$ for all $i$ so that $\prod_{i=1}^{n}{\left(p_i + 1\right)} \equiv \prod_{i=1}^{n}{\left(p_i - 1\right)} \bmod 2$.
We show that it is strictly negative by induction. You prove the base case $n=2$ yourself. Now assume
$$2\prod_{i=1}^{n}{p_i} - \prod_{i=1}^{n}{\left(p_i + 1\right)} - \prod_{i=1}^{n}{\left(p_i - 1\right)} < 0$$
Write this as
$$2R-S-T < 0$$
Then the equivalent expression with $n+1$ terms is
$$2pR - (p+1)S-(p-1)T$$
(where we have written $p$ for $p_{n+1}$). But this is just
$$p(2R-S-T)-S+T$$
which is less than $p(2R-S-T)$ (and therefore $< 0$) because $S=\prod_{i=1}^{n}{\left(p_i + 1\right)}$ is clearly greater than $T=\prod_{i=1}^{n}{\left(p_i - 1\right)}$.
