Let $f\in A(\Omega)$ be an analytic function on a bounded domain $\Omega \subset \mathbb C$ and let $a_0,a_1\cdots a_m\in \omega\Subset \Omega$ where $\omega$ is a smooth open set compactly contained in $\Omega$. Show that $$p(z) = f(z)- \frac{1}{2i\pi}\int_{\partial \omega}\prod_{j=0}^{m}\left(\frac{z-a_j}{\zeta-a_j}\right)\frac{f(\zeta)d\zeta}{\zeta-z}$$
is the unique polynomial of degree at most m that interpolates $f$ at $a_0,a_1\cdots a_m\in \omega $ i.e such that $f(a_j)=P(a_j)$, $j=0,1,\dots ,m .$
My problem is that I am not able to prove that the above defined $P(z)$ is a polynomial. But the fact $f(a_j)=P(a_j)$ follows immediately from Cauchy formula.