I am trying to understand the math. The inverse $\mathcal Z$-transform is given by:
$$x[n] = \displaystyle\frac{1}{j2\pi} \int_cX(z)z^{n-1}dz$$
where $\displaystyle \int_c$ is a contour integral. The inverse Fourier transform is given by:
$$x[n] = \displaystyle\frac{1}{2\pi}\int_{-\pi}^\pi X(e^{j\omega})e^{j\omega n}d\omega$$
My textbook claims, and I agree, that evaluating the inverse z-transform at $z=e^{j\omega}$ will result in the inverse Fourier transform. However, I can't get the math to show this. Substituting $z=e^{j\omega}$, I get as far as:
$$x[n] = \displaystyle \frac{1}{j2\pi}\int_cX(e^{j\omega})e^{j\omega n}e^{-j\omega}d(e^{j\omega})$$
Can somebody tell me intuitively how to simplify this equation to look like the inverse Fourier transform? Is my confusion due to a lack of understanding of the contour integral?