<s>[INST] $n$-derivative of function $f(x)=e^x \sin x$ at $x=0$ I have function $f(x) = e^x \sin{x}$ and must found $f^{(n)}(0)$
$f'(x) = e^x(\sin{x} + \cos{x}) $
$f''(x) = 2 e^x \cos{x}$
$f'''(x) = 2 e^x (\cos{x} - \sin{x})$
$f''''(x) = -4 e^x \sin{x}$
$f'''''(x) = -4 e^x (\sin{x} + \cos{x})$
I think $f^{(n)}(0) = \alpha (-1)^n x^{2n+1}$ but I can't find $\alpha$
 [/INST] I  think the   most   appropriate approach is already given by @JackyChong. 

Slightly more elaborated we have
  \begin{align*}
\color{blue}{\left.\frac{d^n}{dx^n}\left(e^x\sin x\right)\right|_{x=0}}
&=\left.\frac{d^n}{dx^n}\left(\Im e^{(1+i)x}\right)\right|_{x=0}\\
&=\Im \left.\left(\frac{d^n}{dx^n}e^{(1+i)x}\right)\right|_{x=0}\\
&=\Im\left.\left((1+i)^ne^{(1+i)x}\right)\right|_{x=0}\\
&=\Im(1+i)^n\\
&=\Im\left(\sqrt{2}\left(\cos \frac{\pi}{4}+i\sin\frac{\pi}{4}\right)^n\right)\\
&=2^{\frac{n}{2}}\Im\left(\cos\frac{n\pi}{4}+i\sin\frac{n\pi}{4}\right)\\
&\color{blue}{=2^{\frac{n}{2}}\sin\frac{n\pi}{4}}
\end{align*}

</s>