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Prove It $\displaystyle \int_0^{2\pi}{e^{-i(k + n)\theta}\,d\theta} = -\frac{1}{i(k + n)}\left[e^{-i(k + n)\theta}\right]_0^{2\pi}$
$\displaystyle = -\frac{1}{i(k + n)}\left[e^{-2\pi i(k + n)} - e^{0i(k + n)}\right]$.
Assuming that $\displaystyle -(k + n)$ is an integer, these two terms are actually equal (since an integer multiple of $\displaystyle 2\pi$ represents a number of revolutions about the unit circle back to an angle of $\displaystyle 0$).
$\displaystyle = -\frac{1}{i(k + n)}\left[1 - 1\right]$
$\displaystyle = -\frac{1}{i(k + n)}\left[0\right]$
$\displaystyle = 0$.