Maclaurin series of $f(x) = e^x \sin x$ $$f(x) = e^x \sin x$$
I tried applying the given formula in my book but it didn't work.
The maclaurin for $e^x$ is given as $\displaystyle \sum \frac{x^n}{n!}$ and $\sin x$ $\displaystyle \sum \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
I attempted to multiply them together, failed teh books answer. I tried inputting values for them and then multiplying and that fails as well. Why?
$\displaystyle  \frac{x^n}{n!}  \frac{(-1)^n x^{2n + 1}}{(2n+1)!}$
Fails

Since you are looking for the Maclaurin Series of $ f(x):=\exp(x)\sin(x) $ you can start by looking at $f(0)$ and higher derivatives of $f$ at $x=0$. Let $f^{(n)}$ denote the $n$-th derivative of $f$:
$$
\begin{alignat}{2}
f(0) &= \exp(0)\sin(0) &&{}= 0 \\
f^{(1)}(0) &= \exp(0)\sin(0)+ \exp(0)\cos(0) &&{}= 1 \\
f^{(2)}(0) &= 2\exp(0)\cos(0) &&{}= 2 \\
f^{(3)}(0) &= 2\exp(0)\cos(0) - 2\exp(0)\sin(0) &&{}= 2 \\
f^{(4)}(0) &= -4\exp(0)\sin(0) &&{}= 0 \\
f^{(5)}(0) &= -4\exp(0)\sin(0) - 4\exp(0)\cos(0) &&{}= -4 \\
f^{(6)}(0) &=  -8\exp(0)\cos(0) &&{}= -8 \\
f^{(6)}(0) &=  -8\exp(0)\cos(0) + 8\exp(0)\sin(0) &&{}= -8 \\
\dots
\end{alignat}
$$
One can now identify the pattern: Define a real sequence $(a_n)_{n \in \mathbb{N}_0}$ with terms $a_n:= f^{(n)}(0)$. If one now consults the On-Line Encyclopedia of Integer Sequences, one can find an explicit, real, non-recursive expression for former sequence, namely:
$a_n = 2^{\frac{n}{2}}\sin(\frac{n\pi}{4})$
Knowing this, one can write the Maclaurin Series as follows:
$$\begin{align}
\exp(x)\sin(x) 
&= \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n\\
&= \sum_{n=0}^\infty \frac{a_n}{n!}x^n \\
&= \sum_{n=0}^\infty \frac{2^{\frac{n}{2}}\sin(\frac{n\pi}{4})}{n!}x^n \\
&= x + x^2 + \frac{1}{3}x^3 - \frac{1}{30}x^5 - \frac{1}{90}x^6 - \frac{1}{630}x^7  + \frac{1}{22680}x^9 + \frac{1}{113400}x^{10} + \dots \\
\end{align}
$$
