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 $(-1)^{n-1}\rm{Tr}(A)$.