$e^{1/z}$ and Laurent expansion $e^\frac1z$ is not holomorphic at $z=0$, but it is known that it can be expanded as
$$e^\frac1z=1+\frac1z+\frac1{2!z^2}+\frac1{3!z^3}+\cdots$$
The coefficients of this Laurent expansion are computed the same way as Taylor's. The question is how is that possible? If function is not holomoprhic at $z=0$, then it's not true that it is holomophic at $|z|<R$ and Taylor's coefficients can not be used. 
Please someone explain.
 A: I'm pretty sure most of former answers didn't get the key point.
The Laurent series expansion is defined on a "deleted neighborhood" around a singularity, in this case, $\{z: 0<\lvert z-0\rvert< R \}$. In this deleted neighborhood, $e^{1/z}$ is analytic. So for any point in this neighbourhood, we can expand $e^{z}$ first and then substitute $1/z$ in. 
As you can see in the Laurent expansion you gave, you can plug in any $z$ arbitrarily close to zero, calculate the infinite sum, and get a finite and well defined value. The laurent series doesn't give any information about the behavior of a function "on" its singularities.
A: If $f$ is an entire function then
$$
f(z)=a_0 +a_1z+a_2z^2+\ldots.
$$
From Cauchy's integral formula
$a_k=\frac{1}{2 \pi i} \oint_\gamma\frac{f(z)}{z^{k+1}}dz, \ k\in \mathbb{N}$ where $\gamma$ is the unit circle centered at zero.
This means that for all $z\in\mathbb C$
$$
f(z)=a_0 +a_1z+a_2z^2+\ldots.
$$
In particular the above equality holds for all non-zero complex numbers.
In our case the function $f(z)=e^z$ is entire with
$$
f(z)=\sum_{n=0}^\infty\frac{z^n}{n!}.
$$
Since the above equality holds for all non-zero complex numbers it follows that
$$
e^{\frac{1}{z}}=f\left(\frac{1}{z}\right)=\sum_{n=0}^\infty\frac{1}{n!z^n},\qquad \forall z\in\mathbb C.
$$
The above formula for (the Laurent series for) $f(1/z)$ was derived from the Taylor series of $f(z)$ by substituting $1/z$ for $z$ since the Taylor series formula holds for all non-zero complex numbers.
