Showing $\frac{1}{2\pi i}\int_{\vert \gamma \vert =R}f(\zeta) \frac{1-\frac{Q(z)}{Q(\zeta)}}{\zeta - z}d\zeta$ is a Polynomial 
Let $z_1,\ldots, z_n$ be distinct complex numbers contained in the disk $\vert z\vert <R$. Let $f$ be analytic on the closed disk $\vert z \vert \le R$. Let $Q(z)=(z-z_1)\cdots (z-z_n).$ Prove that $$P(z)=\frac{1}{2\pi i}\int_{\vert \gamma \vert =R}f(\zeta) \frac{1-\frac{Q(z)}{Q(\zeta)}}{\zeta - z}d\zeta$$ is a polynomial of degree $n-1$ having the same values as $f$ at the points $z_1,\ldots, z_n$.

I'm not exactly sure how to show that $P$ is holomorphic, but I beleive $P$ extends to a holomorphic on $\vert z \vert <R$ since the poles of the  integrand are removable. Thus we have $P(z)=\sum_{i=0}^\infty \frac{P^{(i)}(0)}{i!}z^i$, so it suffices to show that $P^{(m)}(0)=0$ when $m\ge n$. We have by Cauchy's Integral Formula $P^{(m)}(0)=\frac{1}{2\pi i}\int\frac{f(\zeta)}{\zeta ^{m+1}}d\zeta -\frac{1}{2\pi i}\int \frac{f(\zeta)Q(0)/Q(\zeta)}{\zeta ^{m+1}}$ and both terms are $0$ by the Cauchy Integral Theorem (for the second integral, note that we have removed the poles since $Q(0)=(-1)^nz_1\cdots z_n$)
Am I on the right track? 
 A: The integrand has singularities at $\zeta=z,z_1,z_2,\dotsc,z_m$.
Suppose first that $z \neq z_i$. Then $\frac{Q(\zeta)-Q(z)}{\zeta-z}$ has a removable singularity at $\zeta=z$ because $Q(\zeta) \to Q(z)$ as $\zeta \to z$, so the simple pole cancels with a zero (of order $\geqslant 1$, but this is not relevant yet), and is otherwise analytic. On the other hand, $1/Q(\zeta)$ has simple poles at $\zeta=z_i$, so the Residue Theorem says
$$ \frac{1}{2\pi i} \int_{|\zeta|=R} f(\zeta) \frac{Q(\zeta)-Q(z)}{\zeta-z} \frac{d\zeta}{Q(\zeta)} = \sum_{i} \operatorname{Res}_{\zeta=z_i} f(\zeta) \frac{Q(\zeta)-Q(z)}{\zeta-z} \frac{1}{Q(\zeta)} $$
All the poles are simple, so it suffices to multiply this by $(\zeta-z_i)$ and take the limit.
$$ \operatorname{Res}_{\zeta=z_i} f(\zeta) \frac{Q(\zeta)-Q(z)}{\zeta-z} \frac{1}{Q(\zeta)} = \lim_{\zeta \to z_i} f(\zeta) \frac{Q(\zeta)-Q(z)}{\zeta-z} \frac{\zeta-z_i}{Q(\zeta)} $$
The last factor just results in the product
$$ \prod_{i \neq j} \frac{1}{z_i-z_j}. $$
The first factor is $f(z_i)$. The last thing to demonstrate here is that
$$ \frac{Q(z_i)-Q(z)}{z_i-z} $$
is a polynomial of degree $n-1$. This is easy when we remember that $Q(z_i)=0$, and $z_i-z$ is a factor of $Q$, so what we actually end up with is
$$ \sum_i f(z_i) \prod_{i \neq j} \frac{z_i-z}{z_i-z_j}. \tag{1} $$
It's easy to see this polynomial has the properties we want.
Now, the thing I skated over was $z=z_i$. What happens here? Well, obviously $Q(z_i)=0$, so the integrand reduces to
$$ \frac{f(\zeta)}{\zeta-z_i}, $$
which we know all about: integrating it round the circle gives the value of $f(z_i)$. Hence the function in (1) is actually continuous at $z=z_i$, and really is a polynomial, with no singularities.
Remark: this polynomial is the Lagrange interpolation polynomial for $f$ with base points $z_i$.

The relative simplicity of the above suggests to me that that is the proof desired. On the other hand, to do it your way requires evaluating
$$ \left. \frac{\partial^m}{\partial z^m} \frac{Q(\zeta)-Q(z)}{\zeta-z} \right|_{z=0} = m!\frac{Q(\zeta)}{\zeta^{m+1}} - \sum_{k=0}^m \binom{m}{k} Q^{(k)}(0) \frac{(m-k)!}{\zeta^{m-k+1}} $$
before you integrate, which I can't say I fancy much.
