In the context of the Discrete Fourier Transform, a function in time may be expanded as a Fourier series,
$$ f_m = \frac{1}{T} \sum_{n=0}^{N-1} \tilde{f}_n e^{2\pi inm/N}, $$
where $ f_m $ is the function in the time domain in time step $ m\Delta t $, $ f_m = f(m\Delta t) $, $ T = N \Delta t $ is the period, $ f(m\Delta t) = f(m\Delta t + T) $ and $ \tilde{f}_n $ are the frequency-resolved weights.
The last, are computed in terms of the time-resolved function as
$$ \sum_{m=M}^{M+N-1} f_m e^{-2\pi in'm/N} \Delta t = \frac{1}{N} \sum_{n=0}^{N-1} \tilde{f}_n e^{2\pi i(n-n')M/N} \sum_{m=0}^{N-1} e^{2\pi i(n-n')m/N} = \tilde{f}_{n'}, $$
for any given integer $ M $.
Assuming the above is correct, $ \underline{\text{my question is the following}} $: If we can get the same coefficient $ \tilde{f}_n $ for ANY given integer $ M $, we get that
$$ \sum_{m=M}^{M+N-1} f_m e^{-2\pi inm/N} \Delta t = \sum_{m=0}^{N-1} f_m e^{-2\pi inm/N} \Delta t $$
$$ \Rightarrow \left( e^{-2\pi inM/N} - 1 \right) \sum_{m=0}^{N-1} f_m e^{-2\pi inm/N} = 0 $$
$$ \Rightarrow Mn = \mathbb{Z} N \ \lor \ \tilde{f}_n = 0, $$
which it seems to me obviously wrong since $ M $ can be any integer.
What am I missing?
Your last step is the error. You can't subtract the two sums and assume the values are the same.
If we change the variable in the first sum to $p=m-M,$ you get:
$$\sum_{m=M}^{N+M-1} f_m e^{-2\pi mni/N}=\sum_{p=0}^{N-1}f_{p+M}e^{-2\pi (p+M)ni/N}$$ Now we can subtract this sum from the right sum term by term.
Your result uses $f_p$ where you should have $f_{p+M}.$ You substituted correctly in the exponent, but failed to substitute in the subscript.