As I'm beginning to study the theory of several complex variables, a lot of the early proofs that I've seen from the textbooks I'm using (Range's Holomorphic Functions and Integral Representations in Several Complex Variables, for example) tend to be very cursory because it is claimed that such proofs are easy generalisations of the theory from one variable. I personally don't really like such sketch proofs, so have tried my best to flesh them out in as much detail as possible. The result I'm currently trying to prove is essentially Montel's theorem, and the way that I want to do it is by adapting the proof I know in one variable. This involves showing that if my sequence $(f_j)$ of holomorphic functions is uniformly bounded on every compact subset of a region $D$ of $\mathbb{C}^n$, then it is equicontinuous on every compact subset of $D$, and then by Arzela-Ascoli and a diagonalisation argument, I should be able to extract a compactly convergent subsequence.
The equicontinuity step is what is bothering me. By Cauchy's integral formula, I want to be able to estimate $|f_j(z) - f_j(w)|$ to as small as my liking for every $j$ and for every $z, w$ in a given compact set with $|z - w| < r$. In one variable, this is quite straightforward by taking a slightly smaller disc in the compact set, so that we'd be able to bound $$\left|\frac{1}{\zeta - z} - \frac{1}{\zeta - w}\right| = \frac{|z - w|}{|\zeta - z||\zeta - w|}$$ to our liking, where $\zeta$ lies on the boundary of the larger disc. But for $n$-dimensions, I get something like $$\left|\frac{1}{\prod_{j = 1}^n (\zeta_j - z_j)} - \frac{1}{\prod_{j = 1}^n (\zeta_j - w_j)}\right|,$$ where $\zeta = (\zeta_1, \dots, \zeta_n)$ lies on the distinguished boundary of a polydisc.
My question is how can we perform a similar type of bound in higher dimensions? Cross multiplying the terms do not cancel down to something as nice as $|z - w|$ in one variable anymore, so I'm struggling to see how I can choose an $r > 0$ so that $|z - w| < r$ implies the above is as small as we want.