$n$-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 $\alpha$

You have already found a pattern:
$f^{(1)}(x) = e^x(\sin{x} + \cos{x}) $
$f^{(2)}(x) = 2 e^x \cos{x}$
$f^{(3)}(x) = 2 e^x (\cos{x} - \sin{x})$
$f^{(4)}(x) = -4 e^x \sin{x}$
$f^{(5)}(x) = -4 e^x (\sin{x} + \cos{x})$
This means that $f^{(5)}(x) = -4 f^{(1)}(x)$, $f^{(6)}(x) = -4 f^{(2)}(x)$ and so on.
You can now easily find $f^{(4i)}(x)$, $f^{(4i+1)}(x)$, $f^{(4i+2)}(x)$ and $f^{(4i+3)}(x)$ for all $i \geq 0$ – use induction in $i$. Moreover, since $e^0 = 1$ and $\sin 0 = 0$ and $\cos 0 = 1$, you can now easily find the values of these derivatives at $0$.
