In Alan Oppenheim's book Signals and Systems a comparison is made between the properties of discrete-time and continuous-time complex exponential signals in section 1.3 pg. 26. Specifically it says:
The continuous-time complex exponential $e^{j\omega_0t}$ has two properties: 1. the larger the magnitude of $\omega_0$, the higher is the rate of oscillation in the signal 2. $e^{j\omega_0t}$ is periodic for any value of $\omega_0$
Consider the discrete-time signal $e^{j(\omega_o+2\pi)n}= e^{j2\pi n} e^{j\omega_0 n}=e^{j\omega_0 n}$. Because of the periodicity implied by this equation, the signal $e^{j\omega_0 n}$ does not have a continually increasing rate of oscillation as $\omega_0$ is increased in magnitude. Rather, as illustrated in Figure 1.27, as we increase $\omega_0$ from 0, we obtain signals that oscillate more and more rapidly until we reach $\omega_0=\pi$. As we continue to increase $\omega_0$, we decrease the rate of oscillation until we reach $\omega_0=2\pi$, which produces the same constant sequence as $\omega_0=0$
My confusion is with the part that is in bold. For the continuous-time case, we know that
$\omega_0 = \frac{2\pi}{T_0}$ and it makes sense to me that decreasing $T_0$ (more frequent oscillations) would cause $\omega_0$ to increase. Similarly, for the discrete-time case we know that
$\omega_0 N = 2\pi m$ for $N,m \in \mathbb{Z}$ Solving this equation for $\omega_0$, we get
$\omega_0=\frac{2\pi m}{N}$. Just like in the continuous-time case, decreasing N should cause $\omega_0$ to increase. So I am not seeing what is so special about $\omega_0=\pi$, the rate of oscillations suddenly start decreasing. I don't see that in the math if you decrease N.