= Solution
Choose
$$
\boxed{\chi_A(z)=\det(A-zI)}.
$$
The <Leibniz formula for determinants> shows that its leading term is $(-z)^n$, as required. A scalar $\lambda$ is an <eigenvalue> exactly when $A-\lambda I$ has a nonzero kernel, equivalently when it is singular, so
$$
\boxed{\lambda\text{ is an eigenvalue of }A\iff\chi_A(\lambda)=0}.
$$
The <fundamental theorem of algebra> gives a complex root of the degree-$n$ polynomial $\chi_A$, so every nonempty complex square matrix has at least one complex eigenvalue.
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