I have to evaluate the contour integral $\int_{\gamma} \frac{\sum_{k=0}^m a_k z^k dz}{z^{n+1}} = \int_{\gamma} \frac{f(z)dz}{z^{n+1}}$ with $n \geq 0$ and $\gamma = C(0,1)$ counterclockwise oriented.
So by the generalized Cauchy theorem, we got $$\frac{f^{(n)}(0) 2 \pi i}{n!}=\int_{\gamma} \frac{f(z)dz}{z^{n+1}}$$ and $$\frac{\partial ^{(n)(\sum_{k=0}^m a_k z^k)}}{\partial z^n} = \sum_{k=0}^m a_k \frac{\partial ^{(n)(z^k)}}{\partial z^n}.$$
$$\implies \frac{\partial ^{(n)(z^k)}}{\partial z^n} = k(k-1) \dots (k-n)z^{k-n-1} $$
$$\implies \sum_{k=0}^m a_k \frac{\partial ^{(n)(z^k)}}{\partial z^n} |_{z=0} = a_{n+1}$$
$$\implies \int_{\gamma} \frac{f(z)dz}{z^{n+1}} = \begin{cases} 0 & \quad \text{if } n \geq m\\ \frac{a_{n+1} 2 \pi i}{n!} & \quad \text{if } n < m\\ \end{cases} $$
Does someone could tell me if what I did is good? Otherwise, what can I correct to make the argument valid?
You have $f = \sum a_k z^k$. When you take the $n$th derivative, the sum looks like $n! a_n + C_{n+1} a_{n+1} z + \cdots$. Setting $z = 0$ gives $n! a_n$, not $a_{n+1}$. So then $$ \frac{n! a_n}{n!} = a_n = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z^{n+1}}dz.$$
You can imagine this in a simple way. For all $k \neq n$, the term $$ \frac{1}{2\pi i} \int_\gamma \frac{a_k z^k}{z^{n+1}} dz = \frac{1}{2\pi i} \int_\gamma \frac{a_k}{z^{n+1-k}}dz = 0,$$ as $\frac{1}{z^{b}}$ has a primitive as long as $b \neq 1$. So every term in the integral vanishes, except for the term when $k = n$.
For the one nonvanishing integral, you have $$ \frac{1}{2\pi i} \int_\gamma \frac{a_n z^n}{z^{n+1}}dz = \frac{1}{2\pi i} \int_\gamma \frac{a_n}{z} dz = a_n.$$ And in fact, this is a proof of Cauchy's theorem for derivatives of a function given that analytic functions have power series expansions.