$n$-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 $\alpha$

Observe
\begin{align}
f(x)=e^x\sin x = \operatorname{Im} [e^{(1+i)x}]. 
\end{align}
Using Taylor series for the exponential function, we have
\begin{align}
\operatorname{Im}e^{(1+i)x} = \operatorname{Im} \sum^\infty_{n=0}\frac{(1+i)^nx^n}{n!}.
\end{align}
Thus, it follows
\begin{align}
f^{(n)}(0) = \operatorname{Im}\ (1+i)^n = \operatorname{Im} (\sqrt{2}e^{i\frac{\pi}{4}})^n.
\end{align}
It should be elementary from here on. 
