DFT conjugate of $X^*[k]$, how to prove its formula in terms of $x^*[n]$? Trying to prove that:

$$
X^{*}[k] = \sum^{N-1}_{n=0} x^{*}\left((N-n)\right)_N\ W_N^{nk}
$$

Where: 
$$((x))_N \text{ = x  modulus N}$$

$$W_{N}^{nk} = e^{-j\ 2\pi / N}$$

So I start out with definition for DFT:

$$
X[k]=\sum^{N-1}_{n=0} x[n]\ W_{N}^{nk}
$$

Then I conjugate both sides of equation.

$$
X^{*}[k]=\left[\sum^{N-1}_{n=0} x[n]\ W_{N}^{nk}\right]^{*}
$$

$$
X^{*}[k]=\sum^{N-1}_{n=0} x^{*}[n]\ W_{N}^{-nk}
$$

then somehow the -n changes to N-n which i'm guessing is this:

$$((-n))_N = N - n$$

I don't really understand it... but, seems its equivalent if you do this inside of an exponent… to just randomly take a modulus of a negative signed number and replace it with its modulus equivalent?

$$
X^{*}[k]=\sum^{N-1}_{n=0} x^{*}[n]\ W_{N}^{(N-n)k}
$$

Then, I get stuck on the proof because somehow $x^{*}[n]$ changes to $x^*((N-n))$, 

and $W_{N}^{(N-n)k}$ somehow changes to $W_{N}^{nk}$.  

and thus, i'm unable to obtain final result of:

$$
X^{*}[k] = \sum^{N-1}_{n=0} x^{*}((N-n))_N\ W_N^{nk}
$$

still wondering what the rules are to apply to get there?
So you want to show $X^*[k]$ ?

Actually it follows very simply from the fundamental properties


$x[n] \longleftrightarrow X[k]$
$x[-n] \longleftrightarrow X[-k]$
$x[n]^* \longleftrightarrow X^*[-k]$


and by combining the last two you have:


$x[-n]^* \longleftrightarrow X^*[k]$


which shows that $X^*[k]$ is given by the forward DFT of the signal $x[-n]^*$. 

Note that, due to periodicity of DFT sequences, the negative index can be repaced by $-n = N-n$, and hence $x[-n]^* = x[N-n]^*$ too.

Direct derivation is as follows:

$$ X[k] = \sum_{n=0}^{N-1} x[n] W_N^{nk} $$

where $W_N= e^{-j \frac{2\pi}{N} }$.

Take conjugate of both sides:
$$ X^*[k] = (\sum_{n=0}^{N-1} x[n] W_N^{nk})^* $$

$$ X^*[k] = \sum_{n=0}^{N-1} x^*[n] W_N^{-nk} $$

replace $n$ with $-n$ so that forward DFT appaers:

$$ X^*[k] = \sum_{n=0}^{N-1} x^*[-n] W_N^{nk} $$

which again shows that $X^*[k]$ is given by DFT of $x[-n]^*$. Note again that periodicity of $x[n]$ is utilized to rearrange the limits.