How do I show that $\int_\gamma e^{z+1/z}dz = 2\pi i \sum_{n = 0}^\infty \frac{1}{(n!)(n+1)!} $?

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I'm not sure how exactly to approach this problem. I believe it has something to do with the Laurent Series of the integrand based on the factorial terms in the summation, but I'm not sure exactly how to get there.

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$$\exp(z+1/z)=\sum_{n=0}^\infty\frac{z^n}{n!}\sum_{m=0}^\infty\frac{z^{-m}}{m!} =\sum_{m,n=0}^\infty\frac{z^{n-m}}{n!m!}.$$ If one needed to find the residue at $z=0$ you need to pull out the coefficient of $z^{-1}$, so look at the terms where $m=n+1$.