Trace and the coefficients of the characteristic polynomial of a matrix

The determinant is a sum of (signature-weighted) products of $n$ elements, where no two elements share the same row or column index. From this, it follows, that there is no term with $(n-1)$ terms on the diagonal (if $n-1$ terms of a product are on the diagonal, then the last one must be too, because all other rows and columns are taken). So... the only term that can possibly include a power of $\lambda^{n-1}$ is the product of the main diagonal. Therefore, the $\lambda^{n-1}$ coefficient of $\det A$ equals the $\lambda^{n-1}$ coefficient of $\prod_i (\lambda-A_{ii})$ for which it's easy to show, the coefficient equals $-\sum_i A_{ii}$.

This definition doesn't make assumptions about the field over which the matrix is defined, because the field operations + and * are used directly (with no assumptions about inverses and distribution laws).

Suppose the eigenvalues of $A$ are $\lambda_1,\ldots,\lambda_n$. Then the factored form of the characteristic polynomial is $$(x-\lambda_1)(x-\lambda_2)\cdots(x-\lambda_n).$$

Try using this with induction on $n$.