I am Reading Complex Variables and Applications by Brown and Churchill. On page 230, the author defines the residue of the function $f$ at the isolated singular point $z_0$ as the coefficient of $1/(z-z_0)$ in the Laurent series expansion of $f$. The author then derives the following formula:
$$\int_C f(z)dz=2\pi i\underset{z=z_0}{\operatorname{Res}}f(z)$$
Simple enough. However, the author then gives the following example:
$$\int_C \frac{e^z-1}{z^4}dz=2\pi i\underset{z=0}{\operatorname{Res}}\left(\frac{e^z-1}{z^5}\right)$$
where $C$ is the positively oriented unit circle. Notice that the $z^4$ in the integral has become a $z^5$ in the residue. The author does not mention this and even calculates the residue correctly using $z^5$. Is this a mistake, or is there something going on here that I should be aware of? I am new to this stuff, so it is hard for me to tell what is correct and what is not.
We have for $n\geq 1$, \begin{align} \int_C\frac{e^z-1}{z^n}\,dz &= 2\pi i\cdot \text{Res}\left(\frac{e^z-1}{z^n}; 0\right)=\frac{2\pi i}{(n-1)!}. \end{align} So, one of the sides has a typo.