As to "why take the trace and not any other coefficient of the characteristic polynomial", note that for completely elementary reasons the trace <b>of the whole representation</b> 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.