What is the Laurent series of $ \exp \! \bigl( - \frac{1}{z} \bigr) $? I’m thinking that I could simply let $ x = - \dfrac{1}{z} $ in the Maclaurin series for $ e^{x} $:
$$
  1 - x + \frac{x^{2}}{2!} - \frac{x^{3}}{3!} + \cdots
= 1 - \frac{1}{z} + \frac{1}{2! z^{2}} - \frac{1}{3! z^{3}} + \cdots.
$$
Is that right?

The Taylor series for $e^x$ is 
$$
e^x = \sum_{n=0}^{\infty}\frac{x^n}{n!} = 1 + x + \frac{x^2}{2} + \cdots
$$
Now as you have said, let $x = -\frac{1}{z}$. Then we have
$$
e^{-1/z} = 1 - \frac{1}{z} + \frac{1}{2z^2} - \frac{1}{3!z^3} + \cdots = \sum_{n=0}^{\infty}\frac{1}{(-z)^nn!} = \sum_{n=-\infty}^0\frac{(-z)^n}{(-n)!}
$$
where the Laurent series is 
$$
\sum_{n=-\infty}^0\frac{(-z)^n}{(-n)!}
$$
since to be a Laurent series we need a principal part $\sum_{n=-\infty}^{-1}$ and an analytic part $\sum_{n=0}^{\infty}$
