Does $\frac{1}{z^n}$ have a primitive on $\mathbb{D} - 0$ for $n < 1$?

Question

I'm studying complex analysis from Stein and Shakarachi. There is a question that asks you to evaluate: $\int_{\gamma} z^{n}\ \mathrm{d}z$ for all integer n, where $\gamma$ is any circle centered about the origin. Clearly $n \geq 0 \Rightarrow$ there exists a primitive, which implies the integral around any closed loop is 0. For $n = -1$ the integral is $2\pi i$. For $n < 1$ the integral appears to be 0, since we get: \begin{equation} \begin{split} \int_{\gamma} z^{n}\ \mathrm{d}z & = \int_{0}^{2\pi} (re^{it})^{n} (rie^{it})\ \mathrm{d}t\\ & = \int_{0}^{2\pi} ir^{n+1} e^{it(n+1)}\ \mathrm{d}t\\ & = \frac{r^{n+1}}{n+1} \times (e^{2\pi i(n+1)} - 1) = 0 \end{split} \end{equation} Since $e^{2\pi i n}$ = 1. What's wrong with the calculation? I'm pretty sure $z^{n}$ for $n < -1$ shouldn't have a primitive since they all have a singularity at $z = 0$. Thank you for your help.