For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues. Reading a linear algebra textbook, I encounter

For odd $n$, $n\times n$ matrix with real entries have at least one real eigenvalues.

I noticed in Determinant-free proof that a real $n \times n$ matrix has at least one real eigenvalue when $n$ is odd., the proof without determinant are posts.
It seems for me with the trick of determinant, this problem might be easy.  The first thing I came up with is $\det(A_n-\lambda I_n)$ but this does not guarantee that the solution has at least one real eigenvalues. i.e., setting characteristic polynomial as $f_{A}(\lambda) =\lambda^n + a_{n-1} \lambda^{n-2} + \cdots + a_0$ with $a_i \in \mathbb{R}$, we do not know one of $\lambda$ should be real.
What can be the simple proof with determinants?

The characteristic polynomial is of odd degree.  Since complex roots come in conjugate pairs, and by the fundamental theorem of algebra there are an odd number of roots over $\Bbb C$ (counting multiplicity), there must be a real root.
