Answer
$$e^{4 x}=\sum_{n=0}^{\infty} \frac{(4 x)^{n}}{n !}=\sum_{n=0}^{\infty} \frac{4^{n}}{n !} x^{n}$$
Work Step by Step
Given
$$ f(x)=e^{4x}$$
first we find for $e^x$ , since
\begin{align*}
f(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ f(0)=1\\
f'(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ \ f'(0)=1\\
f''(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ f''(0)=1\\
f'''(x)&=e^x\ \ \ \ \ \ \ \ \ \ \ \ f'''(0)=1\\
f^{(4)}(x)&=e^x \ \ \ \ \ \ \ \ \ \ \ \ \ f^{(4)}(0)= 1
\end{align*}
Then
\begin{align*}
f(x) &=f(a)+\dfrac{f'(a)}{1!}(x-a)+\dfrac{f''(a)}{2!}(x-a)^2+\dfrac{f'''(a)}{3!}(x-a)^3+.....+\dfrac{f^{(n)}(a)}{n!}(x-a)^n\\
&=f(0)+\dfrac{f'(0)}{1!}(x)+\dfrac{f''(0)}{2!}(x)^2+\dfrac{f'''(0)}{3!}(x)^3+.....+\dfrac{f^{(n)}(0)}{n!}(x)^n\\
&= 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots \frac{x^n}{n!}
\\
&=\sum_{n=0}^{\infty}\frac{x^n}{n!}
\end{align*}
To find the Taylor series for $f(x)=e^{4x}$, replace $x$ by $4x$, we get
$$e^{4 x}=\sum_{n=0}^{\infty} \frac{(4 x)^{n}}{n !}=\sum_{n=0}^{\infty} \frac{4^{n}}{n !} x^{n}$$
