Discrete Fourier Transform real f_j's

Question

Could you help me show that if $$\hat{f}(k)=\frac{1}{N}\sum\limits_{j=0}^{N-1}f_j \exp\left(-i\frac{2\pi jk}{N}\right)$$ (k=0,1,...,N-1) is the Discrete Fourier Transform of $f_0, f_1,\ldots, f_{N-1}$, then if the $f_j$'s are real, then $$\widehat{f}(k)=\overline{\widehat{f}(N-k)}$$

Answer 1

$\hat{f}(N-k)= \frac{1}{N} \sum_{j=0}^{N-1}f_j exp(-i \frac{2 \pi j (N-k)}{N})$.

Now it holds $exp(-i 2 \pi j)=1$ due to $e^{2 \pi i}=1$. Take the complex conjugate by using the rule $\bar{ab}=\bar{a} \bar{b}$.