Answer
$$F\left( s \right) = \frac{1}{{s - a}},{\text{ }}s > a$$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( t \right) = {e^{at}} \cr
& {\text{Find the Laplace Transform of the function }} \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}f\left( t \right)} dt, \cr
& {\text{Substitute }}f\left( t \right) = {e^{at}} \cr
& F\left( s \right) = \int_0^\infty {{e^{ - st}}{e^{at}}} dt \cr
& {\text{Recall that }}{e^m}{e^n} = {e^{m + n}} \cr
& F\left( s \right) = \int_0^\infty {{e^{at - st}}} dt \cr
& F\left( s \right) = \int_0^\infty {{e^{\left( {a - s} \right)t}}} dt \cr
& {\text{By the Definition of Improper Integrals with Infinite }} \cr
& {\text{Integration Limits}} \cr
& \int_a^\infty {f\left( x \right)dx = \mathop {\lim }\limits_{b \to \infty } } \int_a^b {f\left( x \right)} dx,{\text{ then}} \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \int_0^b {{e^{\left( {a - s} \right)t}}} dt \cr
& {\text{Integrate}} \cr
& F\left( s \right) = \mathop {\lim }\limits_{b \to \infty } \left[ {\frac{1}{{a - s}}{e^{\left( {a - s} \right)t}}} \right]_0^b \cr
& F\left( s \right) = \frac{1}{{a - s}}\mathop {\lim }\limits_{b \to \infty } \left[ {{e^{\left( {a - s} \right)b}} - {e^{\left( {a - s} \right)\left( 0 \right)}}} \right] \cr
& {\text{Evaluate the limit when }}b \to \infty ,{\text{ }}s > a \cr
& F\left( s \right) = \frac{1}{{a - s}}\left[ {{e^{ - \infty }} - {e^0}} \right] \cr
& F\left( s \right) = \frac{1}{{a - s}}\left( {0 - 1} \right) \cr
& F\left( s \right) = \frac{1}{{s - a}},{\text{ }}s > a \cr} $$
