Laurent series of $e^{1/z}$ centred at $z=1$ If it were $e^z$ being expanded at $z=1$, it would be relatively easy. We would take advantage of the fact that
$$e^z =\sum_{-\infty}^\infty a_n(z-1)^n$$
where $a_n$ is given by the closed-loop integral
$$a_n = \frac{1}{2\pi i} \int_\gamma \frac{e^z}{(z-1)^{n+1}}dz$$
If $n+1\leq0$, then the function inside of the integral is completely defined, so $a_n=0$. So we only worry about when $n \geq 0$. For this, we take advantage of Cauchy's integral formula to get
$$a_n = \frac{e}{n!}$$
And hence
$$e^z = \sum_{n=0}^\infty e(z-1)^n$$
at $z=1$

I wanted to apply the same system to evaluating $e^{1/z}$ at $z=1$. But unfortunately, $\frac{d}{dz}e^{1/z} \neq e^{1/z}$, so the Cauchy integral formula will not be of use. Is there any trick for evaluating this one?
 A: Remark.  A few terms of the two series:
For $|z-1|<1$,
$$
e^{1/z} =
 {\rm e}-{\rm e} \left( z-1 \right) +{\frac {3\,{\rm e}}{2}} \left( z-
1 \right) ^{2}-{\frac {13\,{\rm e}}{6}} \left( z-1 \right) ^{3}+{
\frac {73\,{\rm e}}{24}} \left( z-1 \right) ^{4}-{\frac {167\,{\rm e}
}{40}} \left( z-1 \right) ^{5}+O \left(  \left( z-1 \right) ^{6}
 \right)
$$
For $|z-1|>1$,
$$
e^{1/z} =
1+ \frac{1}{\left( z-1 \right)}-{\frac {1}{2\, \left( z-1 \right) ^{2}}}+{
\frac {1}{6\, \left( z-1 \right) ^{3}}}+{\frac {1}{24\, \left( z-1
 \right) ^{4}}}-{\frac {19}{120\, \left( z-1 \right) ^{5}}}+O \left( 
 \left( z-1 \right) ^{-6} \right) 
$$
A: Let $0<r<1$ and $\gamma$ be the counterclockwise contour $|z-1|=r$.  Then we have for $n\ge 1$ and $|z-1|<1$
$$\begin{align}
a_n&=\frac1{2\pi i}\oint_{\gamma}\frac{e^{1/z}}{(z-1)^{n+1}}\,dz\\\\
&\overbrace{=}^{z=1/w}\frac{(-1)^n}{2\pi i}\oint_{\gamma'}\frac{w^{n-1}e^w}{(w-1)^{n+1}}\,dw\\\\
&=\frac{(-1)^n}{n!} \lim_{w\to 1}\frac{d^n(w^{n-1}e^w)}{dw^n}\\\\
&=\frac{(-1)^n}{n!} \lim_{w\to 1}\sum_{k=0}^n \binom{n}{k}\frac{d^{n-k}e^w}{dw^{n-k}}\frac{d^kw^{n-1}}{dw^k}\\\\
&=\frac{(-1)^n e}{n!}\sum_{k=0}^n \binom{n}{k}\frac{(n-1)!}{(n-1-k)!}\\\\
&=(-1)^n (n-1)! e \sum_{k=0}^{n-1}\frac{1}{k!(n-k)!(n-1-k)!}
\end{align}$$
A: Concerning the first series given by @GEdgar, we have
$$e^{\frac{1}{z}}=e\sum_{n=0}^\infty a_n\,(z-1)^n$$ where the $a_n$ are given by
$$n a_n+(2 n-1) a_{n-1}+(n-2) a_{n-2}=0$$ with $a_0=1$ and $a_1=-1$
In absolute value, the numerators form sequence $A067764$ in $OEIS$; denominators form sequence $A067653$.
In terms of special functions
$$a_n=\frac {U(-n,0,-1)}{n!}$$
$U(a,b,z)$ being known as the Kummer's function of the second kind, Tricomi function, or Gordon function. This is a concise form of @Mark Viola's formula.
