Analytic function and the Lagrange interpolating polynomial

Let $f(z)$ be analytic in a closed set $A$ and bounded by contour $C$. Let $z_1,z_2,...,z_n$ be different arbitrary point in the interior of $C$ and let $w_n(z)=(z-z_1)(z-z_2)...(z-z_n) $. Prove the integral $$P(z)=\frac{1}{2\pi i} \int_C \frac{f(x)}{w_n(x)} \frac{w_n(x)-w_n(z)}{x-z} dx$$ is a polynomial of degree $(n-1)$ that is the same as $f(z)$ at $z_1,z_2,...,z_n$ points.

Note: P(z) is the Lagrange interpolating polynomial.

I think I have an idea for the proof, but I don't know how to write it formally.

Idea I think this formula $$f^n(z_0)=\frac{n!}{2\pi i} \int_J \frac{f(x)}{(x-z_0)^{n+1}} dx $$ will be useful because practically is telling us that P$(z)$ must be 'equal to' $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$.

Now the rest is trying to proof that $\frac{f^n(z_0)}{n!} \left(\frac{w_n(x)-w_n(z)}{x-z} \right)$ must be something very similar to the Lagrange polynomial.

Is this a good idea? Could be demonstrated in another way? Thanks for any suggestions/hint/tip/help.

Part 1 Using Cauchy's integral formula $$P(z_k)=\frac{1}{2\pi i}\int_{C}\frac{f(x)}{w_n(x)}\frac{w_n(x)-w_n(z_k)}{x-z_k}dx=\frac{1}{2\pi i}\int_{C}\frac{f(x)}{w_n(x)}\frac{w_n(x)}{x-z_k}dx=\frac{1}{2\pi i}\int_{C}\frac{f(x)}{x-z_k}dx=f(z_k)$$ for $\forall k=1..n$.

Part 2 Same Cauchy's integral formula $$P(z)=\frac{1}{2\pi i}\int_{C}\frac{f(x)}{w_n(x)}\frac{w_n(x)-w_n(z)}{x-z}dx=\frac{1}{2\pi i}\int_{C}\frac{f(x)}{x-z}\left(1-\frac{w_n(z)}{w_n(x)}\right)dx=\\ \frac{1}{2\pi i}\int_{C}\frac{f(x)}{x-z}dx-\frac{1}{2\pi i}\int_{C}\frac{f(x)}{x-z}\frac{w_n(z)}{w_n(x)}dx=\\f(z)-\frac{w_n(z)}{2\pi i}\int_{C}\frac{f(x)}{x-z}\frac{1}{w_n(x)}dx$$ or $$f(z)-P(z)=\frac{w_n(z)}{2\pi i}\int_{C}\frac{f(x)}{(x-z)w_n(x)}dx \tag{1}$$

Proposition 1 $$\frac{1}{w_n(z)}=\sum_{j=1}^{n}\frac{1}{(z-z_j)w_{n}'(z_j)} \tag{2}$$

Proof Polynomial $$q(x)=\sum_{j=1}^{n}\frac{w_n(z)}{(z-z_j)w_{n}'(z_j)}-1$$ is of degree $n-1$ and has $n$ zeros, since $q(z_k)=0,\forall k=1..n$, thus $q(x)\equiv 0$.

From $(1)$ $$\frac{w_n(z)}{2\pi i}\int_{C}\frac{f(x)}{(x-z)w_n(x)}dx=\frac{w_n(z)}{2\pi i}\int_{C}\frac{f(x)}{(x-z)}\left( \sum_{j=1}^{n}\frac{1}{(x-z_j)w_{n}'(z_j)} \right)dx=\\ \frac{w_n(z)}{2\pi i} \left( \sum_{j=1}^{n} \frac{1}{w_{n}'(z_j)}\int_{C}\frac{f(x)}{(x-z)(x-z_j)} dx \right)=\\ \frac{w_n(z)}{2\pi i} \left( \sum_{j=1}^{n} \frac{1}{w_{n}'(z_j)(z-z_j)}\int_{C}\frac{f(x)(z-z_j)}{(x-z)(x-z_j)} dx \right)=\\ \frac{w_n(z)}{2\pi i} \left( \sum_{j=1}^{n} \frac{1}{w_{n}'(z_j)(z-z_j)}\left[\int_{C}\frac{f(x)}{(x-z)} dx -\int_{C}\frac{f(x)}{(x-z_j)} dx \right]\right)=\\ \frac{w_n(z)}{2\pi i} \left( \sum_{j=1}^{n} \frac{2\pi i}{w_{n}'(z_j)(z-z_j)}\left[f(z) -f(z_j) \right]\right)=\\ w_n(z) \left( \sum_{j=1}^{n} \frac{f(z) -f(z_j)}{w_{n}'(z_j)(z-z_j)}\right)=w_n(z) \left( \sum_{j=1}^{n} \frac{f(z)}{w_{n}'(z_j)(z-z_j)} - \sum_{j=1}^{n} \frac{f(z_j)}{w_{n}'(z_j)(z-z_j)} \right)=...$$ from $(2)$ $$...=f(z) - \sum_{j=1}^{n} \frac{w_n(z)f(z_j)}{w_{n}'(z_j)(z-z_j)}$$ from $(1)$ $$f(z)-P(z)=f(z) - \sum_{j=1}^{n} \frac{w_n(z)f(z_j)}{w_{n}'(z_j)(z-z_j)} \Rightarrow P(z) = \sum_{j=1}^{n} \frac{w_n(z)f(z_j)}{w_{n}'(z_j)(z-z_j)}$$ which makes $P(z)$ a polynomial of degree $n-1$.

Some references here.