What is the Laurent series of $e^{z+z^{-1}}$? (Detailed derivation using Cauchy product of series)

Question

We know that: $$e^{z+z^{-1}}=\underbrace{\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\cdot\left(\sum_{m=0}^{\infty}\frac{z^{-m}}{m!}\right)}_{=:\alpha}.$$ Now I am having trouble, applying the following (Cauchy product for series) on $\alpha$: $$\left(\sum_{i=0}^{\infty}a_iz^i\right)\cdot\left(\sum_{j=0}^{\infty}b_jz^j\right)=\sum_{k=0}^{\infty}c_k z^k, \quad c_k=\sum_{l=0}^{k}a_lb_{k-l}. \quad (\ast)$$ My approach was the following:
With ($\ast$) I find that \begin{align} \left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\cdot\left(\sum_{m=0}^{\infty}\frac{z^{-m}}{m!}\right) &= \sum_{k=0}^{\infty}c_k, \\ c_k &= \sum_{l=0}^{k}\left(\frac{1}{l!}\frac{1}{(-(k-l))!}\right) \\ &= \sum_{l=0}^{k}\frac{1}{l!(l-k)!}, \end{align} which results in $$e^{z+z^{-1}}=\sum_{k=0}^{\infty}\left(\sum_{l=0}^{k}\frac{1}{l!(l-k)!}\right)z^k.$$ The result I am looking for is $$e^{z+z^{-1}}=\sum_{k=-\infty}^{\infty}\left(\sum_{l=0}^{\infty}\frac{1}{l!(l+k)!}\right)z^k, \quad \forall k\in\mathbb{N}_0.$$ What am I missing? Also, I suspect that I have to use the following $$\alpha=\left(\sum_{n=0}^{\infty}\frac{z^n}{n!}\right)\cdot\left(\sum_{m=-\infty}^{0}\frac{z^m}{(-m)!}\right)$$ in order to get a sum from $-\infty$ to $\infty$, but I do not know how.