Laplace Transform of Exponential/Real Argument

Theorem
Let $\laptrans f$ denote the Laplace transform of a function $f$.

Let $e^x$ be the real exponential.

Then:
 * $\map {\laptrans {e^{a t} } } s = \dfrac 1 {s - a}$

where $a \in \R$ is constant, and $\map \Re s > \map \Re a$.