This can easily be verified by substituting the definition of the DFT into the IDFT definition, changing the summation index of the DFT from n to m to avoid name capture problems (n is a dummy variable, so can be changed freely without altering the value of the expression):
The inner summation above is zero for all m<>n, and equal to N for m=n, (m,n = 0..N-1). (The proof of this is presented in Annex B.) In other words, using 'Kroneker Delta' notation:
So, the expression for f(n) becomes:
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