I am learning about the discrete Fourier transform in the context of digital image processing and not in the context of complex analysis so I might have some slightly different notation/formulas.
Our professor has told us that in the 1D discrete case:
$$\mathcal{F}(1) = \delta(u)$$
if you use the formula:
$$\mathcal{F}(f(t)) = \sum_{t=0}^{M-1}f(t)*e^{-i 2 \pi ut/M}$$
Where $\delta(u)$ is the 1D discrete impulse function defined as:
$$ \delta(u) = \left\{ \begin{array}{lr} 1, & \text{if } u =0\\ 0, & \text{if } u\neq 0 \end{array} \right\} $$
I understand that the inverse discrete Fourier transform of $\delta(u)$ is $f(t)=1$ but I seem to be having trouble showing it the other way around. Here is what I have so far.
$$\mathcal{F}(1) = \sum_{t=0}^{M-1} e^{-i2\pi ut/M} = \sum_{t=0}^{M-1} (e^{-i2\pi })^{ut/M}$$
If $u \neq 0$ then each term is equal to 0 because $e^{-i2 \pi}$ is always 0 by Euler's identity.
If $u=0$ then each term in the sum is just 1 so the total sum is equivalent to $M$ instead of $1$. What am I doing wrong here?