It is well known that the infinite product $\prod (1+a_n)$ converges if the sum $\sum a_n$ converges, for $a_n>0$.
The proof uses $\exp(x)\leq 1+x$ for $x>0$.
$$ \prod \left( 1+a_n \right) \leq \prod \left( e^{a_n} \right) = e^{\sum a_n} $$
If the sum converges, so does the product, because the RHS $\geq$ LHS.
Question: If we consider $a_n$ to be complex, does the following inequality prove a similar rule: $\prod (1+a_n)$ converges if the sum $\sum |a_n|$ converges?
$$ \left| \prod \left( 1+a_n \right) \right | \leq \prod \left( 1+|a_n| \right) \leq \prod \left( e^{|a_n|} \right) = e^{\sum |a_n|} $$
Is this is a sufficient and correct proof?
Note: Some sources add a condition that $a_n\neq-1$, but I don't see why because the sum would diverge anyway. What am i missing?
Edit: added abs bars as per comments below.
$A_n:= \prod_{k=1}^n (1+a_k)$
$B_n:= \prod_{k=1}^n (1+|a_k|)$
So $|A_n| \le |B_n|$. Now, for all $n<m$: $$ \begin{align} |A_n-A_m|&=|A_n|\left| \prod_{k=n+1}^m (1+a_k)-1\right| \\ &=|A_n| \left| \sum_{i=1}^{m-n} \sum_{E \subset \{n+1,\dots,m\} : \#E=i} \prod_{j \in E} a_j \right| (*) \\ &\le |A_n|\left| \sum_{i=1}^{m-n} \sum_{E \subset \{n+1,\dots,m\} : \#E=i} \prod_{j \in E} |a_j|\right| \\ &=|A_n|\left| \prod_{k=n+1}^m (1+|a_k|)-1\right| \\ &\le |B_n| \left| \prod_{k=n+1}^m (1+|a_k|)-1\right| \\ &=|B_m-B_n| \end{align}$$ where the "mysterious sum" in (*) is just a formal way to write the expansion $$( 1+b_1)(1+b_2)...(1+b_l)- 1=(b_1+\dots)+(b_1b_2+\dots)+\dots$$ So $|A_n-A_m| \le |B_n-B_m|$. Besides $B_n$ converges, hence the conclusion.