Answer
$\Sigma_{k=0}^{n} \dfrac {e(x-1)^k}{k!} $
Work Step by Step
The nth Taylor polynomial for a function $f$ is
$$p_x(x)= f(x_0) + f'(x_0)(x-x_0) +\dfrac{f''(x_0)}{2!} (x-x_0)^2 +.......+\dfrac{f^n(x_0)}{n!} (x-x_0)^n$$
$f(x)=e^x$
Differentiate w.r.t x
$f'(x)=e^x$
$f''(x)=e^x$
$f'''(x)=e^x$
$f''''(x)=e^x$
Now $f(0)=e$ , $f'(0)=e$ , $f''(0)=e$, $f'''(0)=e$, $f''''(0)=e$
Hence the Maclaurin Series for the function is
$\displaystyle e+e(x-1)+\dfrac{e(x-1)^2}{2} .......+ \frac {e(x-1)^n}{n!} =\Sigma_{k=0}^{n} \frac {e(x-1)^k}{k!} $
