proof of Autocorrelation property of DFT I am facing problem in proving the Auto-correlation property of Discrete Fourier Transform (DFT), that is

$$\mathcal{DFT}\left\{\sum_{r=0}^{N-1}x[r]x^*[r+n]\right\} = X[k]X^*[k]= |X[k]|^2$$

where $X[k]$ is the DFT of $x[n]$, and $x[n]$ is periodic with period $N$:
$$ x[n+N] = x[n] \qquad \forall n \in \mathbb{Z} $$
Likewise  $X[k]$ is periodic with period $N$:
$$ X[k+N] = X[k] \qquad \forall k \in \mathbb{Z} $$

Below is my Approach:

$$\begin{align}
\mathcal{DFT}\left\{\sum_{r=0}^{N-1}x[r]x^*[r+n]\right\} &amp;= \sum_{n=0}^{N-1} \sum_{r=0}^{N-1}x[r] x^*[r+n] e^{- j 2 \pi n k/N} \\ \\
 &amp;= \sum_{r=0}^{N-1}x[r] \sum_{n=0}^{N-1}  x^*[r+n] e^{- j 2 \pi n k/N} \\ \\
 &amp;= \sum_{r=0}^{N-1}x[r] \sum_{n=r}^{r+N-1}  x^*[n] e^{- j 2 \pi (n-r) k/N} \\ \\
 &amp;= \sum_{r=0}^{N-1}x[r] \sum_{n=0}^{N-1}  x^*[n] e^{j 2 \pi r k/N} e^{- j 2 \pi n k/N} \\ \\
 &amp;= \sum_{r=0}^{N-1}x[r]  e^{j 2 \pi r k/N}  \sum_{n=0}^{N-1}  x^*[n] e^{- j 2 \pi n k/N} \\ \\
 &amp;= \sum_{r=0}^{N-1}x[r]  e^{j 2 \pi r k/N}  \sum_{n=0}^{N-1}  x^*[n] (e^{- j 2 \pi n (-k)/N})^* \\ \\
 &amp;= \sum_{r=0}^{N-1}x[r]  e^{j 2 \pi r k/N}  \sum_{n=0}^{N-1} ( x[n] e^{- j 2 \pi n (-k)/N})^* \\ \\
 &amp;= \sum_{r=0}^{N-1} x[r] e^{ j 2 \pi r k/N} X^*[-k] \\ \\
 &amp;= \sum_{r=0}^{N-1} x[r] e^{ -j 2 \pi r (-k)/N} X^*[-k] \\ \\
 &amp;= X[-k] X^*[-k] = \Big| X[-k] \Big|^2 \\ \\
 &amp; \ne \ X[k] X^*[k] \quad = \Big| X[k] \Big|^2 \\
\end{align}$$

as $$X[k] = \mathcal{DFT} \{ x[r] \} = \sum_{r=0}^{N-1} x[r] e^{- j 2 \pi r k/N}$$

so where is my mistake? and how should I approach??
You did not consider the conjugate property of the DFT.

The complex conjugate property of DFT states that
$\rm{DFT}\{x^{*}[n]\} = X^{*}[-k]$. 
This implies that exponent in the second line of your proof should be $-k$. Which then allows for the DFT of $x[n]$ as $X[k]$. 
A number multiplied by its conjugate give the square of its absolute value. 

And DFT is discrete in the transform domain, hence its parameter should be in brackets rather than parentheses.