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