I am following these notes on signals and systems. Following is the paragraph from page 25 which I am having a difficulty with.
Discrete-time sinusoids have the obvious form $s(n) = A\cos(2πfn + φ)$. As opposed to analog complex exponentials and sinusoids that can have their frequencies be any real value, frequencies of their discrete-time counterparts yield unique waveforms only when $f$ lies in the interval $(-1/2, 1/2]$ This property can be easily understood by noting that adding an integer to the frequency of the discrete-time complex exponential has no effect on the signal’s value.
$$e^{i 2 \pi (f+m) n}=e^{i 2 \pi f n}e^{i 2 \pi mn}=e^{i 2 \pi f n} $$
This derivation follows because the complex exponential evaluated at an integer multiple of $2π$ equals one.
I don't understand how can $e^{i 2 \pi mn }$ be always $1$.
Suppose $f'=f+m=0.75$ where $-1/2<f \le 1/2$,
then $f=1/2$ and $m=0.25$ but $e^{i 2 \pi 0.75 n}\neq1 \forall n \in \mathbb{Z}$
The point of the paragraph is to show that replacing $f$ with $f+m$ does not change the waveform, for any integer m.
This would imply that it is only relevant to choose values of $f$ in an interval of length $1$, since going any wider would result in duplicates. The author of the text chooses the interval $(-\frac12, \frac12]$: you can see that there is no way to get more than one point of this set, if we only add integers; moreover, for any real number we may choose, by adding an integer we can always get to one (and only one) point of this set.
To prove this, we must show that this relation holds: $$ e^{i2\pi (f+m) n}=e^{i2\pi f n}$$
Starting from the left side, simply distribute the exponent: $$ e^{i2\pi (f+m) n}=e^{i2\pi f n+i2\pi m n}$$
Then, apply a property of the exponential function: $$e^{i2\pi f n+i2\pi m n}=e^{i2\pi f n}e^{i2\pi m n}$$
Since we chose $m$ as an integer, and $n$ is also integer, then $2\pi m n$ is an integer multiple of $2\pi$, and therefore $e^{i2\pi m n}=e^0=1$.
Thus we have proved our claim.