TITLE: Show that: $p_{n+1}\nmid2^{p_1p_2...p_n}+1$
QUESTION [5 upvotes]: Let $p_1>p_2>...>p_{n+1}>3$ prime numbers. Show that: $$p_{n+1}\nmid2^{p_1p_2...p_n}+1$$
I tried to prove it by contradiction using Fermat's Theorem but i'm struggling.

REPLY [2 votes]: Below put $\,p= p_{n+1},\ k = p_1\cdots p_n,\ a = 2$ 
Lemma $\ $ Prime $\,p\mid a^k+1\,\Rightarrow\, p\mid a^2-1\ $ if $\ p\nmid a\,$ and $\,\color{#c00}{(k,p\!-\!1)=1}.\ $ Proof:
$\!\bmod p\!:\ (a^2)^k\equiv 1\equiv (a^2)^{p-1}\!\Rightarrow\, a^2\equiv 1\,$ by $\,a^2\,$ has order $\,j\!=\!1$, by $\,j\mid \color{#c00}{k,p\!-\!1\ \rm{coprimes}}$