In Digital Filter Design by Parks and Burrus, p. 19.
The transfer function of an FIR filter is given by the $\mathcal Z$-transform of $h(n)$ as:
$$H(z)=\sum_{n=0}^{N-1}h(n)z^{-n}$$
(where $h$ is the filter)
The frequency response of a filter is defined as
$$H(\omega)=\sum_{n=0}^{N-1} h(n)e^{-j\omega n}$$
where $\omega$ is frequency in $\textrm{rad/sec}$.
Then the text proceeds to show that $H(\omega)$ is periodic with period $2\pi$:
\begin{align} H(\omega + 2 \pi)&= \sum_{n=0}^{N-1} h(n) e^{-j(\omega+2\pi)n}\\ &= \sum_{n=0}^{N-1} h(n) e^{- j \omega n} \color{red}{e^{-j2\pi n}}\\ &=H(\omega) \end{align} Could someone clarify how is that equal to $H(\omega)$ when there's the extra term $\color{red}{e^{-j2 \pi n}}$?