How can I prove $\det(\overline M)=\overline{\det(M)}$? Of course $\overline M$ is the complex conjugate of an $n\times n$ matrix $M$.  
Someone gave me advice to use the definition of determinant, then it means I have to use cofactor expasion here?  
 A: You have the following definition of the determinant:
$$\det(M) = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}m_{i,\sigma(i)}.$$
Then one has $$\det(\bar M) = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}\bar m_{i,\sigma(i)} = \sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\overline{\prod_{i=1}^{n}m_{i,\sigma(i)}} = \overline{\sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^{n}m_{i,\sigma(i)}} = \overline{\det M},$$ all this equalities holding because conjugation is a $\mathbb R$-algebra homomorphism.
A: If you want to use the Laplace development (cofactors), then you can do it by induction.
The case for $1\times1$ matrices is obvious, so let $n>1$ and assume the result for $(n-1)\times(n-1)$ matrices.
If $A=[a_{ij}]$ is an $n\times n$ matrix, denote by $A_{ij}$ the matrix obtained by removing the $i$-th row and the $j$-th column. To not complicate notations, let $B=\bar{A}=[b_{ij}]$ with $b_{ij}=\overline{a_{ij}}$.
The expansion of $\det B$ along its first line is
$$\det B=(-1)^{1+1}b_{11}\det B_{11}+(-1)^{1+2}b_{12}\det B_{12}+\dots+
(-1)^{1+n}b_{1n}\det B_{1n}$$
By induction hypothesis, $\det B_{ij}=\overline{\det A_{ij}}$, so you can write
\begin{align}
\det B&=
(-1)^{1+1}\overline{a_{11}}\,\,\overline{\det A_{11}}+
(-1)^{1+2}\overline{a_{12}}\,\,\overline{\det A_{12}}+\dots+
(-1)^{1+n}\overline{a_{1n}}\,\,\overline{\det A_{1n}}
\\[4pt]
&=
(-1)^{1+1}\overline{a_{11}\det A_{11}}+
(-1)^{1+2}\overline{a_{12}\det A_{12}}+\dots+
(-1)^{1+n}\overline{a_{1n}\det A_{1n}}
\\[4pt]
&=\overline{
  (-1)^{1+1}a_{11}\det A_{12}+
(-1)^{1+2}a_{12}\det A_{12}+\dots+
(-1)^{1+n}a_{11}\det A_{1n}
}
\\[4pt]
&=\overline{\det A}
\end{align}
