Laplace Transform of Exponential/Real Argument

Theorem
Let $e^x$ be the real exponential.

Let $\mathcal L$ be the Laplace transform.

Then:
 * $\displaystyle \mathcal L \left\{ {e^{a t} }\right\} \left({s}\right) = \dfrac 1 {s - a}$

where $a \in \R$ is constant, and $\operatorname{Re}\left({s}\right) > \operatorname{Re}\left({a}\right)$.