I'm going to define the characteristic polynomial to my students next monday, and I'm going to tell them about its degree and three special coefficients. The constant coefficient is trivial by evaluation at zero. For both others and the degree, the reasoning is an example of what you ask, if I got your question right.

I will write the determinant on the board (written with parenthesis because I don't remember how to make the straight lines...):
$$\begin{pmatrix}a_{1,1}-X & a_{1,2} & \dots \\\
a_{2,1} & a_{2,2}-X & \dots \\\
\dots & \dots & \dots
\end{pmatrix}$$
and consider the development along the first column : the first term is $a_{1,1}-X$ times a cofactor which looks the same as the original determinant, and all others will have lost *two* terms of the form $\text{something}-X$ so will at least two lower in degree. That means if I were to really develop the determinant, it would end up looking like $(a_{1,1}-X)\dots(a_{n,n}-X)+\text{at most degree}(n-2)$, so the characteristic polynomial is degree $n$, its dominant coefficient is $(-1)^n$, and the coefficient just behind will be $(-1)^{n-1}\rm{Tr}(A)$.