$\displaystyle \mathcal{L} \left\{f(t)\right\} = \int_0^\infty e^{-st} f(t) \, dt$
$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-st} e^{at} \, dt$
$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-st + at} \, dt$
$\displaystyle \mathcal{L} (e^{at}) = \int_0^\infty e^{-(s - a)t} \, dt$
$\displaystyle \mathcal{L} (e^{at}) = -\dfrac{1}{s - a} \int_0^\infty e^{-(s - a)t} \, [ \, -(s - a) \, dt \, ]$
$\displaystyle \mathcal{L} (e^{at}) = -\dfrac{1}{s - a} \left[ e^{-(s - a)t} \right]_0^\infty$
$\mathcal{L} (e^{at}) = -\dfrac{1}{s - a} \left[ \dfrac{1}{e^{(s - a)t}} \right]_0^\infty$
$\mathcal{L} (e^{at}) = -\dfrac{1}{s - a} \left[ \dfrac{1}{\infty} - \dfrac{1}{e^0} \right]$
$\mathcal{L} (e^{at}) = -\dfrac{1}{s - a} (0 - 1)$
Thus,
$\mathcal{L} (e^{at}) = \dfrac{1}{s - a}$ answer
