Finding the Taylor series expansion of $f(z)=\frac{e^{z}-1}{z}$ around $0$ 
Find the Taylor series expansion of $f(z)=\displaystyle\frac{e^{z}-1}{z}$ around $0$.

I have no idea where to start.

You know that 
$$\exp(x)=\sum_{k=0}^\infty \frac{x^k}{k!}$$
so 
$$\frac{\exp(z)-1}{z}= \sum_{k=1}^\infty \frac{z^{k-1}}{k!}$$
