Answer
$e \Sigma_{n=0}^{\infty} \dfrac{(x-1)^{n}}{n!}$
Work Step by Step
Here, we have: $f(x)=e^x \implies f(1)=e\\ f'(x)=e^x \implies f'(1)=e\\ f''(x)=e^x \implies f''(1)=e$
and so on.
The Taylor series of a function $f(x)$ at $x=c$ can be written as:
$T(x)=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+.......\dfrac{f^n(a)}{n!}(x-a)^n$
Hence, the required Taylor series is:
$T(x)=e+\dfrac{e}{1!}(x-1)+\dfrac{e}{2!} (x-1)^2+....=e \Sigma_{n=0}^{\infty} \dfrac{(x-1)^{n}}{n!}$
