We have an amazing thing called Lagrange remainders. They basically tell us the difference between our function and it's Taylor polynomial. In general, we have
$$R_n(x)=|f(x)-P_n(x)|$$
where $P_n(x)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(x-a)^k$. Since it follows that
$$R_n(a)=0\\R_n'(a)=0\\R_n''(a)=0\\\vdots\\R_n^{(n)}(a)=0\\R_n^{(n+1)}(a)=|f^{(n+1)}(a)|$$
Thus,
$$R_n^{(n+1)}(x)\le|f^{(n+1)}(c)|$$
for some $c$ in our radius of convergence. It thus follows by integrating a few times that
$$R_n(x)\le\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|$$
One can then see that as $n\to\infty$, we have
$$|f(x)-P(x)|\le\lim_{n\to\infty}\left|\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\right|$$
and if $\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}\to0$ for any $x,c$ within the a given domain, then the power series will equal the original function over that domain.
See if you can show that for any $x,c\in\mathbb R$,
$$\lim_{n\to\infty}\left|\frac{e^c}{(n+1)!}x^{n+1}\right|=0$$
On a side note, Lagrange remainder also shows us how well we approximate something when using a power series. For example, if I wanted to calculate $e$ out 5 places accurately,
$$R_n(x)=\left|e^x-\sum_{k=0}^n\frac{x^n}{n!}\right|\le0.000001$$
It's easy enough to solve, since
$$R_n(x)\le\left|e\frac{x^{n+1}}{(n+1)!}\right|\le\left|3\frac{x^{n+1}}{(n+1)!}\right|$$
Our particular case is $x=1$, and thus it suffices to solve
$$\frac3{(n+1)!}<0.000001$$
Which is easily done with a few checks to give $n\le8$. Thus,
$$e=\pm0.000001+\sum_{k=0}^8\frac1{k!}$$
Best Answer
Just write $e^{2x}(e^{x}-1)$ as $e^{3x}-e^{2x}$. The coefficident is $\frac {3^{24}} {(24)!} -\frac {2^{24}} {(24)!}$.