Trace and determinant via eigenvalues

Leibniz formula.
$$\chi_A(\lambda)=\det(\lambda I-A)=\sum_{\sigma\in S_n}\varepsilon(\sigma)\prod_i(\lambda\delta_{i,\sigma(i)}-a_{i,\sigma(i)}).$$

Leading term. The highest power of $\lambda$ comes from the identity permutation, giving $\lambda^n$.

Coefficient of $\lambda^{n-1}$. The identity permutation contributes $-(\sum_i a_{ii})\lambda^{n-1}=-\operatorname{tr}(A)\lambda^{n-1}$. All other permutations contribute at most $\lambda^{n-2}$ (since they have at most $n-2$ fixed points, hence at most $n-2$ diagonal entries).

Constant term. At $\lambda=0$, $\chi_A(0)=\det(-A)=(-1)^n\det A$.

Conclusion.
$$\chi_A(\lambda)=\lambda^n-\operatorname{tr}(A)\lambda^{n-1}+\cdots+(-1)^n\det A.$$
In particular, the trace and determinant are coefficients (up to sign) of the characteristic polynomial, hence invariant under similarity.