Minimal polynomial for an invertible matrix and its determinant So here's one that I can't quite crack:

Let $A\in M_n(\mathbb{F})$ be an invertible matrix with integer eigenvalues. Its minimal
  polynomial is $m_A(\lambda)=\lambda^k+b_{k-1}\lambda^{k-1}+...+b_0$. Prove that $detA$ is divisible by $b_0$.

I'm afraid I don't quite see the connection between the minimal polynomial and the determinant.
All I know is that for an invertible matrix $b_0 \neq 0$, but I can't find a way to relate that to the determinant. 
Any tips and guidance would be much appreciated!
Thanks in advance.
 A: Hint:
For any square matrix of order $n$, the constant term of the characteristic polynomial $\chi_A$is $(-1)^n\det A$, and the minimal polynomial of $A$ divides $\chi_A$. 
A: There exists a polynomial $p(t)$ such that
$$
\DeclareMathOperator{char}{char}\char_A(t)=p(t)m_A(t)
$$
It follows that
$$
(-1)^n\det(A)=\char_A(0)=p(0)m_A(0)=p(0)b_0
$$
That is, 
$$
\det(A)=(-1)^np(0)b_0
$$
Hence $b_0$ divides $\det(A)$.
A: For any ring $R$, and any polynomials $f(x), d(x), q(x) \in R[x]$, we have
$f(x) = d(x) q(x) \Rightarrow f_0 = d_0 q_0, \tag{1}$
where $f_0$ is the constant term of $f(x)$ etc.  This is easy to see by simply examining the basic rule for polynomial multiplication, with
$d(x) = \sum_0^{\deg d} d_i x^i, \tag{2}$
$q(x) = \sum_0^{\deg q} q_i x^i, \tag{3}$
$f(x) = \sum_0^{\deg d + \deg q} f_j x^j = d(x)q(x) = \sum_{j = 0}^{j = \deg d + \deg q} (\sum_0^j d_i q_{j - i})x^j; \tag{4}$
looking at the coefficient of $x^0$ in the inner sum, we see from (4) that (1) binds.  Now if the eigenvalues $\lambda_i$, $1 \le i \le n$,  of $A$ are integers, it follows that $\det A = \prod_1^n \lambda_i$ is itself an integer, and since we may write the characteristic polynomial $p_A(x)$ as
$p_A(x) = \prod_1^n (x- \lambda_i), \tag{5}$
it follows from the fact that the minimal polynomial $m_A(x)$ of $A$ must divide $p_A(x)$ that $m_A(x)$ may be written
$m_A(x) = \prod_{i \in S} (x - \lambda_i) \tag{6}$
for some subset $S \subset \{1, 2, \ldots, n \}$; thus we see that the degree-zero term of
$m_A(x) = \sum_0^{\vert S \vert} b_j x^j, \tag{7}$
where $\vert S \vert$ is the cardinality of $S$, is given by
$m_0 = \pm \prod_{j \in S} \lambda_j, \tag{8}$
so $m_0$ itself is also an integer.  It then follows from (1) etc. above that $m_0 \mid \det A$, as integers, the desired conclusion.
Nota Bene:  Assuming $\Bbb F$ is a field ($\Bbb F$ for field, right?), the fact that $m_A(x)$ may be factored in the form (6) follows from the fact that $\Bbb F[x]$ is a principal ideal domain, hence unique factorization into primes (which in a PID are essentially the same as irreducibles) applies; the $x - \lambda_i$, each being irreducible in $\Bbb F[x]$, are thus primes etc. etc. etc.  End of Note.
A: All roots of the minimal polynomial are eigenvalues, so here that (monic) polynomial is of the form $\prod_{i=1}^k(X-a_i)$ for some sequence $(a_1,\ldots,a_k)$ of integers, all eigenvalues, with maybe some repetitions. Then $b_0= \prod_{i=1}^k(-a_i)$. For the characteristic polynomial the situation is the same, but maybe with more repetitions of certain eigenvalues. The constant term then is an integer multiple of $b_0$, and on the other hand the constant term of the characteristic polynomial is $\det(-A)$. The result follows.
