Question: Is it true that for any real matrix, the complex eigenvalues occur in complex conjugate pairs? If yes, how does one prove that?

Answer: 
$det(A) = = det \left( \begin{matrix}a\_1 \\ \vdots \\ a\_{n-1}\\a\_n \end{matrix} \right) = det \left( \begin{matrix}a\_1 \\ \vdots \\ a\_{n-1} \\a\_1 + \cdots +a\_n \end{matrix} \right) = det \left( \begin{matrix}a\_1 \\ \vdots \\ a\_{n-1}\\0 \end{matrix} \right) = 0$

Where $a\_i$ are the row vectors of $A$.

Of course, within the matrix $0$ stands for the row vector $0 = (0 \ldots 0)$.
Suppose the matrix is $(a\_{ij}$, you have $a\_{ni}=\sum\_{j=1}^{j=n-1}-a\_{ji}$, if you sum the first $n-1$ line you have the last line with the opposite sign.