nth derivative taylor series

Let $\displaystyle f^{(n)}(a)$ denote the n-th derivative of $\displaystyle f$ at $\displaystyle a$. if $\displaystyle f(x)=xe^{2x}$, then $\displaystyle f^{(64)}(0)=$

I can get the answer using the old fashion way, but when I lookeed at the solution using power series, I don't understand one of the steps. heres the soultion my prof has:

$\displaystyle e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$

$\displaystyle f(x)=xe^{2x}=\sum_{n=0}^\infty \frac{2^nx^{n+1}}{n!}=\sum_{m=1}^\infty \frac{2^{m-1}x^{m}}{(m-1)!}= \boxed{\sum_{m=1}^\infty \frac{f^{(m)}(0)}{m!}x^m}$

hence, $\displaystyle f^{(m)}(0)=\frac{2^{m-1}m!}{(m-1)!}$

I understand every step except the step to the result in the box. How are they equal?