Proof of uniform convergence of an infinite product I am reading the book Complex Analysis: An Invitation (2nd Edition), page 163-164. There is a certain step in the proof, which I can not fill the details. First, I mention a relevant proposition and a definition.
Proposition: The infinite product $\prod_{n=1}^{\infty}(1+a_k)$ converges if $\sum_{n=1}^{\infty}|a_n|<\infty$, and in that case
$$
  \left | \prod_{k=1}^\infty (1+a_k)-1 \right |\leq e^{\sum_{k=1}^\infty |a_k|}-1 \tag{*}
$$
Definition: Let $(a_n)_{n\geq 1}$ be a sequence of complex valued functions defined in an open subset $\Omega$ of $\mathbb{C}$. We say that the infinite product $\prod_{n=1}^{\infty}(1+a_k)$ converges locally uniformly in $\Omega$, if $\prod_{n=1}^{\infty}(1+a_k(z))$ converges at each $z\in \Omega$ and if furthermore to each compact subset $K$ of $\Omega$ and each $\epsilon>0$ there exists an $N$ such that for all $z\in K$ and $n\geq N$
$$
\left | \prod_{k=1}^{\infty}(1+a_k(z))-\prod_{k=1}^{n}(1+a_k(z)) \right |<\epsilon.
$$
What I want to prove is:

Lemma: Let $(a_n)_{n\geq 1}$ be a sequence of complex valued functions on an open subset $\Omega$ of $\mathbb{C}$. If as $N\to\infty$ the sum $\sum_{n=N}^{\infty}|a_n(z)|$ converges locally uniformly to $0$, then the infinite product $\prod_{n=1}^{\infty}(1+a_k)$ converges locally uniformly in $\Omega$.

The proof which the author says is simply: "It is a consequence of Proposition and the inequality (*)" It does not seem completely clear to me. Could someone explain that step for me?
Update: When I read the answer of Kavi Rama Murthy, I thought as follows: Let $K$ be any compact subset of $\Omega$, and let $\epsilon\in (0,1/2)$ be given. Choose an integer $N^*$ with $N^*\geq N$ such that $\sum_{n=n+1}^{\infty}|a_n(z)|<\epsilon$ for all $z\in K$ and all $n\geq N^*$. Then, we have for all $z\in K$ and all $n\geq N^*$
$$
\left | \prod_{k=n+1}^\infty (1+a_k(z))-1 \right |\leq e^{\sum_{k=n+1}^\infty |a_k(z)|}-1<e^\epsilon - 1<2\epsilon
$$
and so
$$
\left | \prod_{k=1}^{\infty}(1+a_k(z))-\prod_{k=1}^{n}(1+a_k(z)) \right |=\left | \prod_{k=1}^{n}(1+a_k(z)) \right |\left | \prod_{k=n+1}^{\infty}(1+a_k(z))-1 \right |<2\epsilon \left | \prod_{k=1}^{n}(1+a_k(z)) \right |.
$$
Then I got stuck here.

$|\prod_{k=n}^{m}(1+a_k(z))-1|\leq e^{ \sum\limits_{k=n}^{m}|a_k(z)}-1$. On any compact set $K$ we can choose $N$ such that $e^{ \sum\limits_{k=n}^{m}|a_k(z)}-1<\epsilon$ for alll $z \in K$ fro all $n,m \geq N$. Now consider $|\prod_{k=1}^{n}(1+a_k(z))-\prod_{k=1}^{m}(1+a_k(z))|$ where $n <m$. We can write this as $|\prod_{k=1}^{n}(1+a_k(z))| |\prod_{k=n+1}^{m}(1+a_k(z))-1|$. What remains is to see that th first factor $\prod_{k=1}^{n}(1+a_k(z))$ is uniformly bounded on $K$. This follows by another application of (*). I hope you can finish.
