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