Does the Cauchy integral formula apply to negative values of $n$? The Cauchy integral formula is as follows:
$$f^{n}(a) = \dfrac{n!}{2 \pi i}\oint_C \dfrac{f(z)}{(z-a)^{n+1}}\,\mathrm{d}z.$$
However in every source I can find describing the Cauchy integral formula does not state the domain of $n$ and I have only ever seen it used for positive values of $n$.  So my question is does the formula hold for negative values of $n$?  If so what does $f^{n}$ mean when $n$ is negative?
 A: The following excerpt from Applied and Computational Complex Analysis, Vol. 1 by P. Henrici might be helpful since it explicitely states the domain.

Theorem 4.7b (Cauchy Integral formula)
Let $R$ be a simply connected domain, let $f$ be analytic everywhere in $R$, and let $z_0$ be a point of $R$. Then, if $\Gamma$ is a piecewise regular closed curve in $R$ not passing through $z_0$,
  \begin{align*}
n(\Gamma,z_0)f(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{z-z_0}\,dz.
\end{align*}
  In particular, if $\Gamma$ is a positively oriented Jordan curve that contains the point $z_0$ in its interior, then
  \begin{align*}
f(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{z-z_0}\,dz.
\end{align*}

and later on

Corollary 4.7c
Under the hypotheses of Theorem 4.7b, if $\color{blue}{k=0,1,2,\ldots}$,
  \begin{align*}
n(\Gamma,z_0)f^{(k)}(z_0)=\frac{k!}{2\pi i}\int_{\Gamma}\frac{f(z)}{(z-z_0)^{k+1}}\,dz,
\end{align*}
  and particularly, if $\Gamma$ is a positively oriented Jordan curve containing $z_0$ in its interior,
  \begin{align*}
\frac{1}{k!}f^{(k)}(z_0)=\frac{1}{2\pi i}\int_{\Gamma}\frac{f(z)}{(z-z_0)^{k+1}}\,dz.
\end{align*}

Conclusion: In the corollary 4.7c the domain of $k$ is explicitely stated to be non-negative integers. We conclude, that the Cauchy Integral formula stated in OPs question does not apply to negative $n$.
A: $\frac{f^{n}(a)}{n!}$ is the coefficient of $(z-a)^n$ in the series expansion of $f(z)$ around the point $z = a$. An analytic function does not have negative powers in the series expansion, therefore we can define this to be zero for an analytic function, while for a meromorphic function we can take such coefficients to be given by the Laurent expansion coefficients. Then the Cauchy integral is indeed zero for analytic functions when you take $n$ negative, while for meromorphic functions that have a pole at $z = a$ and no other poles inside the contour, you'll also get the correct answer.
So, the conclusion is that it does work for negative $n$ when you replace $\frac{f^{n}(a)}{n!}$ by the coefficient of $(z-a)^n$.
