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Laplace Transform: Find the sum of the series $\sum_{n=0}^\infty \frac{e^-s\left(\frac{1}{n!}\right)}{n!}$
Let $A(n) = \frac{e^-s\left(\frac{1}{n!}\right)}{n!}$, how do we find the Laplace transform of this series $\sum_{n=0}^\infty A(n)$?
I’ve tried writing
$$\sum_{n=0}^\infty A(n) = \sum_{n=0}^\infty \frac{e^-s}{n!} = \exp(-s) \sum_{n=0}^\infty \frac{1}{n!} = \exp(-s) \frac{1}{\exp(s)-1} \infty$$
I get $1/(s\exp(s)-s)$ but I am not sure how I get $1$
A:
\begin{align*}
\mathcal{L} \{e^{ -\lambda t}\sum_{n=0}^\infty \frac{\lambda^n t^n}{n!}\}
&=\sum_{n=0}^\infty \frac{\lambda^n \mathcal{L} \{t^n\}}{n!}\\
&=\lambda \sum_{n=0}^\infty \frac{1}{n!}\\
&=e^{ -\lambda}
\end{align
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