Answer
\begin{aligned}
e^{x} &= 1+\frac{1}{1 !} x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\ldots \ldots+\frac{1}{n !} x^{n}
\end{aligned}
Work Step by Step
Since
\begin{aligned} f(x) &=e^{x} & & f(0)=1 \\ f^{\prime}(x) &=e^{x} & & f^{\prime}(0)=1 \\ f^{\prime \prime}(x) &=e^{x} & & f^{\prime \prime}(0) =1 \\ f^{\prime \prime}(x) &=e^{x} & & f^{\prime \prime \prime}(0) =1 \\ f^{(4)}(x) &=e^{x} & & f^{(4)}(0) =1 \end{aligned}
Then
\begin{aligned}
e^{x} &=f(0)+\frac{f^{\prime}(0)}{1 !} x+\frac{f^{\prime \prime}(0)}{2 !} x^{2}+\frac{f^{\prime \prime \prime}(0)}{3 !} x^{3}+\ldots \ldots+\frac{f^{(n)}(0)}{n !} x^{n} \\
&=1+\frac{1}{1 !} x+\frac{1}{2 !} x^{2}+\frac{1}{3 !} x^{3}+\ldots \ldots+\frac{1}{n !} x^{n}
\end{aligned}
