Laurent Series of $\frac{1}{z^2+1}$ around $0<|z|<1$

Question

I have to find the Laurent Series of $$f(z) = \frac{e^z}{z(z^2+1)}$$ which has singularities $z=0$ and $z=\pm i$. I know that $e^z = \sum_{n=0}^{\infty} \frac{z^n}{n!}$ and that $\frac{1}{1-z} = \sum_{n=0}^{\infty}z^n$ for $|z|<1$. In the solutions they say that using these two known series I can find $$\frac{1}{z^2+1} = 1-z^2+z^4-z^6+...$$ but how do I find it from them?