How do I evaluate the following integral?
$$\oint\_{C}\frac{e^z-1}{z}\mathrm dz$$ where $C$ is the unit circle (counter-clockwise).

I have just learned Cauchy-Goursat's Theorem, but I cannot apply it here since there is a singularity at $z=0$. So I try to go brute force and evaluate the contour integral as follows:
$$\int\_{0}^{1}\dfrac{e^{e^{it}}-1}{e^{it}}ie^{it}\,\mathrm dt = \int\_{0}^{1}({e^{e^{it}}-1})i\,\mathrm dt .$$
But then I'm stuck again as to how to evaluate this. 

Thanks!