I am referring to a solved example from Brown and Churchill's Complex variables and applications (ninth ed. page 231). If I simply use the Maclaurin series (which I think is a specific case of Laurent series), shouldn't the integral be $2\pi{i}*\frac{1}{3!}$ because the positive powers in the Laurent series are given by $$a_n = \frac{1}{2\pi{i}}\int_C\frac{f(z)}{z^{n+1}}dz$$ In this case since the denominator is $z^4$ isn't $n=3$ and we can simply use the Maclaurin series expansion of $e^z-1$ and find the $a_3$ term which happens to be $\frac{1}{3!}$ This gives the answer as $\frac{\pi{i}}{3}$ but the answer in the book is $\frac{\pi{i}}{12}$ using the residue for $$\frac{e^z-1}{z^5}$$ at the point $z=0$
I am also not sure why the book calculates residue with the integrand with denominator $z^5$ in the solution. You can also look here at this image:
https://i.stack.imgur.com/SMROJ.png
You are right !
$\oint_C f(z) dz= 2i\pi \sum_{a \in C} Res(f(z), a)$.
Here, $f(z) = \frac{e^z-1}{z^4} = \frac1{z^3} + \frac1{2! z^2} + \frac1{3!z}+ \frac1{4!}+\ldots$
the residue in $z=0$ is $1/3! = 1/6$ so $\oint_C f(z) dz=\frac{2i\pi}{6} = \frac{i\pi}{3}$ and the response of the book is false.