Chapter 9 - Infinite Series - 9.8 Maclaurin And Taylor Series; Power Series - Exercises Set 9.8 - Page 667: 2

$e^{ax}= \Sigma_{k=0}^{\infty} a^k \dfrac{x^k}{k!}$

$e^x=\Sigma_{k=0}^{\infty} \dfrac{x^k}{k!}$ We will replace $x$ with $ax$. $e^{ax}= \Sigma_{k=0}^{\infty} \dfrac{(ax)^k}{k!}$ So, $e^{ax}= \Sigma_{k=0}^{\infty} a^k \dfrac{x^k}{k!}$