Suppose that $f \in H(\Omega)$ and suppose that $\Omega$ contains the closure of some polydiscs $D(p;r)$. By polydiscs I means $D(p;r)=\{z=(z_1,z_2,..,z_n) \in \mathbb{C^n}| |z_i-p_i| \lt r_i \}$ for $i=1,2,..,n$. Here $p=(p_1,p_2,..,p_n) \in \mathbb{C^n}$ and $r=(r_1,r_2,..,r_n)$. I want to show that $f$ has a power series expansion in the polydisc.
I know that there is a cauchy integral formula for $f$ in $D(p,r)$. So for any point $z \in D(p,r)$ I have $$f(z)=\int_{|w-p|=r}\prod_{i=1}^{n}(w_i-z_i)^{-1}f(w) dw$$
Here $w=(w_1,w_2,..,w_n)$ . So I look at $\dfrac{1}{w_i-z_i}=\dfrac{1}{w_i-p_i+p_i-z_i}=\dfrac{1}{\left(w_i-p_i\right) \left(1-\dfrac{z_i-p_i}{w_i-p_i}\right)}$
Now since $|z_i-p_i| \lt |w_i-p_i|=r$, we have $$\dfrac{1}{w_i-z_i}=\frac{1}{w_i-p_i}\sum_{n=0}^{\infty}\left(\dfrac{z_i-p_i}{w_i-p_i}\right)^n=\sum_{n=0}^{\infty}\dfrac{\left(z_i-p_i\right)^n}{\left(w_i-p_i\right)^{n+1}}$$
Thus I can write $$f(z)=\int_{|w-p|=r}\prod_{i=1}^{n}\left(\sum_{n=0}^{\infty}\dfrac{\left(z_i-p_i\right)^n}{\left(w_i-p_i\right)^{n+1}}\right)f(w) dw$$ $$=\int_{|w-p|=r}\sum_{ \alpha \in \mathbb{N^n}}\left(\dfrac{\left(z-p\right)^{\alpha}}{\left(w-p\right)^{\beta}}\right)f(w) dw=\sum_{\alpha \in \mathbb{N^n}}(z-p)^{\alpha} \int_{|w-p|=r}\frac{f(w)}{(w-p)^{\beta}} dw=\sum_{\alpha}\gamma(\alpha)(z-p)^{\alpha}$$
Here $\beta$ depends on $\alpha$ for if $|\beta|=|\alpha|+1$ where $\alpha=(\alpha_1,..,\alpha_n)$ and when I sum over $\alpha$ , it covers $\beta$ as well.
I am a little lost over the interchange of integration and summation, for I know that each such series converges uniformly and $f$ is continuous in a closed polydisc, So I can appeal to Weirstrass-M test . But I am not sure how to conclude that the product of series is convergent or for that matter I want to appeal to the Abel's lemma which says that
$$\text{Suppose that $\{c_{\alpha}\}_{\alpha \in \mathbb{N^n}}$ is a set of complex numbers such that, for some $p_1,p_2,..,p_n \gt 0$, there is a $M \gt 0$ for which $|c_{\alpha}|{p_{1}}^{\alpha_{1}}..{p_{n}}^{\alpha_{n}} \le M$ for all $\alpha \in \mathbb{N^n}$ , then the series $\sum_{\alpha \in \mathbb{N^n}} c_{\alpha}(z-a)^{\alpha} $ converges uniformly for $|z_j-a_j| \le \theta p_j, 0 \le \theta \lt 1$} $$
The proof of this lemma is easy as $|\sum_{\alpha \in \mathbb{N^n}} c_{\alpha}(z-a)^{\alpha}| \le \sum_{n}\theta^{n}$. I am unable to find such a bound for the series as in the proof of Cauchy Integral Formula .
An arbitrary term in the series will look like some $$\gamma_{\alpha}(n)=\sum_{|\alpha|=n}\frac{(z_1-p_1)^{\alpha_{1}}(z_2-p_2)^{\alpha_{2}}..(z_k-p_k)^{\alpha_{k}}..(z_n-p_n)^{\alpha_{n}}}{(w_1-p_1)^{\beta_{1}}(w_2-p_2)^{\beta_{2}}..(w_k-p_k)^{\beta_{k}}..(w_n-p_n)^{\beta_{n}}} $$
where $|\beta|=|\alpha|+1$
Thanks for the help!!