Answer
$\Sigma_{k=0}^{\infty} \dfrac {e (x-1)^k}{k!}$
Work Step by Step
The Maclaurin Series is
$ \displaystyle\sum\limits_{n=0}^{\infty} \frac{f^n(0)x^n} {n!} = f(0) + f'(0)x + \frac{(f''(0)x^2)}{2!} + ..\\
$
$f(x)=e^x$
Differentiate:
$f'(x)=e^x$
$f''(x)=e^x$
$f'''(x)=e^x$
$f''''(x)=e^x$
Now $f(1)=e$ , $f'(1)=e$ , $f''(1)=e$, $f'''(1)=e$, $f''''(1)=e$
Hence the Maclaurin Series for the function is
$\displaystyle e+e(x-1)+ \frac {e} {2}(x-1)^2 + .......+ \frac {e}{n!}(x-1)^n = \Sigma_{k=0}^{\infty} \frac {e (x-1)^k}{k!}$
