Uniform convergent series Let $\Omega$ be domain in $\mathbb C^2$. For each compact set $K_j$ define the holomorphic function $f_j$ on $\Omega$, such that $$\sup_{k_j}|f_j|<2^{-j}.$$
Define  $$f= \prod_{j=1}^\infty(1-f_j)^j.$$
I need to show that this product converges uniformly on each compact set $K_l$.
 A: We use the following result:

Let $S$ a set and $u_n\colon S\to \Bbb C$ functions such that $\sum_{n\geq 0}|u_n(s)|$ is uniformly convergent on $S$. Then the product $\sum_{n=1}^{+\infty}(1+u_n(s))$ is uniformly convergent on $S$. 

We need a lemma:

If $N$ is an integer and $c_1,\ldots,c_N$ are complex numbers, and writing $p_N:=\prod_{j=1}^N(1+c_j)$, $p_N^*=\prod_{j=1}^N(1+|c_j|)$, we have 
  $$p_N^*\leq \exp\left(\sum_{j=1}^N|c_j|\right)\mbox{ and }|p_N-1|\leq p_N^*-1.$$
  The first inequality follows from $1+x\leq e^x$ if $x\geq 0$, and the second con be shown by induction. 

Let $P_n(s):=\prod_{j=1}^n(1+u_j(s))$; we have 
\begin{align}|P_{n+m}(x)-P_n(s)|&=|P_n(s)|\left|\prod_{j=n+1}^{n+m}(1+u_j)-1\right|\\\
&\leq \exp\left(\sum_{j=1}^n|u_j(s)|\right)\left(\prod_{j=n+1}^{n+m}(1+|u_j|)-1\right)\\\
&\leq \exp\left(\sum_{j=1}^{+\infty}|u_j(s)|\right)\left(\exp\left(\sum_{j\geq n+1}|u_j(s)|\right)-1\right).
\end{align}
We choose, for a fixed $\varepsilon>0$ a $n_0$ such that $\sup_{s\in S}\sum_{j\geq n_0+1}|u_j(s)|\leq \varepsilon$, and we get the wanted result. 

Now, we deal with this particular case. Let $u_j:=(1-f_j)^j-1$. we have to prove that $\sum_{j\geq 1}|u_j(s)|$ is uniformly convergent on $K_l$. 
We have for $n\geq l$ that 
\begin{align*}
|u_n(s)|&=\left|\sum_{k=0}^n\binom nk(-f_n(s))^{n-k}-1\right|\\\
&=\left|\sum_{k=1}^n\binom nk(-f_n(s))^{n-k}\right|\\\
&\leq \sum_{k=1}^n\binom nk|f_n(s)|^{n-k}\\\
&= \sum_{k=1}^n\binom nk|f_n(s)|^k\\\
&\leq \sum_{k=1}^n\binom nk 2^{-nk}\\\
&=2^{-n}\sum_{k=0}^{n-1}2^{-kn}\\\
&=2^{—n}+\sum_{k=0}^{n-1}2^{-n(k+1)}\\\
&\leq (n+1)2^{-n},
\end{align*}
which is enough to conclude since $\sum_{n\geq 1}(n+1)2^{-n}$ is converging. 
