Contour integral around circle - Complex Analysis Exercise :

Show that :
$$\frac{1}{2πi}\oint_{|z|=r} \frac{e^{1/z}}{z-1}dz = \Bigg\{\begin{matrix}
1-e & r<1\\ 
1 & r >1
\end{matrix}
$$

I came across this exercise while I was studying for my complex analysis semester exams and I would really appreciate any guidance, tips or thorough solution. I know I am not providing an attempt (probably the only time I have done so in all my questions) but I can't grasp on how to work depending on the radius $R$. Most integrals I've solved can be worked around with the Theorem of Residues or the Trigonometric/General Integrals techniques, but I cannot seem to work on this one. Maybe one idea is to work through poles and singular points.
I would really value a thorough explanation.
 A: An idea for the residues
$$z=0:\;\;\frac{e^{1/z}}{z-1}=-\frac1{1-z}\cdot e^{1/z}=-\sum_{n=0}^\infty z^n\cdot\sum_{n=0}^\infty\frac1{n!z^n}=$$
$$=-\sum_{n=0}^\infty\sum_{k=0}^n\frac{z^k}{(n-k)!z^{n-k}}=-\sum_{n=0}^\infty\sum_{k=0}^n\frac1{(n-k)!z^{n-2k}}$$
The coefficient of $\;z^{-1}\;$ in the above is given by $\;n-2k=1\iff k=\cfrac{n-1}2\;$ , and thus $\;n\;$ must be odd, so we get that the coefficient for each such value of $\;n\;$ odd is $\;\frac1{\left(\frac{n+1}2\right)!}\;$ , and thus the whole wanted coefficient is
$$-\sum_{n=1}^\infty\frac1{n!}=-(e-1)=1-e$$
I think the residue at $\;z=1\;$ is much easier to calculate and I leave that to you.
A: Hint


*

*If $r<1$, the only singularity is at $0$. Since 
$$\frac{1}{z-1}=-\sum_{n\geq 0}z^n,$$
you can conclude easily compute the Laurent series of $\frac{e^{1/z}}{z-1}$ and conclude.

*If $r>1$, you also have a singularity at $z=1$. Since $e^{1/z}$ is holomorphic at the neighborhood of $1$, you can easily compute the residue at $z=1$ using Cauchy formula.
