# Complex Exponential Sequence (Discrete-time exponentials) factor equalling 1?

I'm reading page 10 of Schaum's ouTlines "Signals and Systems" by Hwei P. HSU, and I don't understand why this factor in the complex exponential sequence equals 1.

Consider the complex exponential sequence with frequency ($$\Omega_{0} + 2\pi k$$), where $$k$$ is an integer:

$$e^{j(\Omega_{0} + 2\pi k)n} = e^{j\Omega_{0}n}e^{j2\pi kn} = e^{j\Omega_{0}n}$$ $$\quad$$ $$(1.56)$$

since $$e^{j2 \pi k n} = 1$$.

My question: why does $$e^{j2 \pi k n} = 1$$ which, to my understand, would imply that $$k$$ is not just any integer but $$0$$ for this to be true since $$j$$ can't be $$0$$ and $$n$$ can't be $$0$$ because then $$e^{j\Omega_{0}n}$$ would also equal $$1$$.

The texts proceeds:

From Eq. $$(1.56)$$ we see that the complex exponential sequence at $$\Omega_{0}$$ is the same as that at frequencies ($$\Omega_{0} \pm 2\pi$$), ($$\Omega_{0} \pm 4\pi$$), and so on. Therefore, in dealing with discrete-time exponentials, we need only consider an interval of length $$2\pi$$ in which to choose $$\Omega_{0}$$.

$$\exp(j2\pi kn) = \cos(2\pi kn) + j\sin(2\pi kn)$$, where $$\cos(2\pi kn) = 1$$ for $$kn \in \mathbb{Z}$$, i.e., integer. And $$\sin(2\pi kn) = 0$$ for $$kn \in \mathbb{Z}$$. Thus $$\exp(j2\pi kn) = \cos(2\pi kn) + j\sin(2\pi kn) = 1$$ for $$kn \in \mathbb{Z}$$