In Digital Filter Design by Parks and Burrus, p. 19.

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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 $\color{red}{e^{-j2 \pi n}}$?