I toook very basic example:
$$\oint_\limits{z = |1|} z^3 e^{\frac{1}{z}} \, dz$$
well, it is very well seen that $z = 0$ is the essentional singular point, since that, we have to apply Laurent's series for exponential expansion, after all the steps I got:
$$e^\frac{1}{z} = z^3 + z^2 + \frac{z}{2} + \frac{1}{4!z} + \ldots +$$
I have two questions:
1.) as the series can be expanded to any possible power we want, which power should I take as the greatest? (in the given case and just in general)
2.) which formula should I apply to calculate a residue for an essentional singularity?
As you have already the Laurent series around the pole, it is very easy:
$$z^3e^{1/z}=z^3\left(1+\frac1z+\frac1{2z^2}+\frac1{6z^3}+\frac1{24z^4}+\ldots\right)=\ldots+\frac1{24\,z}+\ldots$$
and thus the residue is $\;\cfrac1{24}\;$
Observe that we need the Laurent series (or power series, depending on the case) up to exponents that render the $\;z^{-1}\;$ summand.