As to "why take the trace and not any other coefficient of the characteristic polynomial", note that for completely elementary reasons the trace of the whole representation still knows the characteristic polynomial of each individual element.
For instance the second-from-top coefficient of the characteristic polynomial of $\rho(g)$ is $\frac{1}{2}(\mathrm{tr}(\rho(g))^2 - \mathrm{tr}(\rho(g^2))).$ Writing down the formula for subsequent coefficients is an exercise with symmetric functions. On the other hand, the higher coefficients of the characteristic polynomial do lose information -- e.g. non-isomorphic representations rather often have the same determinant.
