Answer
$e^x=e^a[1+(x-a)+\dfrac{(x-a)^2}{2!}+...+\dfrac{(x-a)^n}{n!}]$
Work Step by Step
The Taylor series for $f(x)$ at $x=a$ can be written as:
$\Sigma_{n=0}^{\infty} \dfrac{f^{k}(a)}{k!}(x-a)^k=f(a)+f'(a)(x-a)+.....+\dfrac{f^{n}(a)}{n!}(x-a)^n$
This yields: $f(x)=e^x$ and $f^n(a)=e^a$
Now, $e^ x=e^a+e^a(x-a) +\dfrac{1}{2!} e^a (x-a)^2 +......$
Thus, we have the result:
$e^x=e^a[1+(x-a)+\dfrac{(x-a)^2}{2!}+...+\dfrac{(x-a)^n}{n!}]$
