Let A be an nxn matrix with characteristic polynomial
$\displaystyle f(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1}+...+ a_1\lambda +a_0 $
Prove that $\displaystyle a_0 = (-1)^ndetA$
Recall that $\displaystyle f(t)=\det (A-tI)$. Set $\displaystyle t=0$ on both sides! You should just get $\displaystyle a_0$. The factor $\displaystyle (-1)^n$ could come into the picture if somehow you had defined $\displaystyle f(t)=\det (tI-A)$ but that is certainly not standard. The answer is just $\displaystyle a_0$.