For the function defined as follows, find the Taylor polynomials of degree 4 at 0....

The function is, ${\displaystyle f\left(x\right)=e^{x+1}}$
The Taylor polynomials of degree 4 at 0 is,
${\displaystyle P_4\left(x\right)=f\left(0\right)+\frac{f^{\prime }\left(0\right)}{1!}x+\frac{f{}^{″}\left(0\right)}{2!}{x}^2+\frac{f^{\left(3\right)}\left(0\right)}{3!}{x}^3=\frac{f^{\left(4\right)}\left(0\right)}{4!}{x}^4}$
Find the value of $f(x)$ at $x=0$ as,
${\displaystyle f\left(0\right)=e^{0+1}}$
${\displaystyle =e^1}$
$=e$
Find the first, second, third and fourth derivatives of $f(x)$ and find its corresponding values at $x=0$ as shown in below table.
$\begin{array}{|ll|}\hline \text{Derivative of function, }f(x)=e^{x+1}& \text{Value of derivative at }x=0\\ f^{\prime }(x)=e^{x+1}& f^{\prime }(0)=e^{0+1}=e\\ f^{″}(x)=\frac{d}{dx}(f^{\prime }(x))=\frac{d}{dx}(e^{x+1})=e^{x+1}& f^{″}(0)=e^{0+1}=e\\ f^{(3)}(x)=\frac{d}{dx}(f^{″}(x))=\frac{d}{dx}(e^{x+1})=e^{x+1}& f^{(3)}(0)=e^{0+1}=e\\ f^{(4)}(x)=\frac{d}{dx}(f^{(3)}(x))=\frac{d}{dx}(e^{x+1})=e^{x+1}& f^{(4)}(0)=e^{0+1}=e\\ \hline\end{array}$
The Taylor polynomials of degree 4 at 0 is calculated as follows.
${\displaystyle P_4\left(x\right)=f\left(0\right)=\frac{f^{\prime }\left(0\right)}{1!}x+\frac{f{}^{″}\left(0\right)}{2!}{x}^2+\frac{f^{\left(3\right)}\left(0\right)}{3!}{x}^3+\frac{f^{\left(4\right)}\left(0\right)}{4!}{x}^4}$
${\displaystyle =e+\frac{e}{1}x+\frac{e}{2}{x}^2+\frac{e}{6}{x}^3+\frac{e}{24}{x}^4}$