Q:

$A$'s minimal: $M_A(t)=\prod_{i=1}^{k}(t-\lambda_i)^{s_i}$, find $\bigl(\begin{smallmatrix} A &I_n \\ I_n & A \end{smallmatrix}\bigr)$'s minimal 

Let $A \in M_n(\mathbb R)$ and suppose its minimal polynomial is:
$$M_A(t)=\prod_{i=1}^{k}(t-\lambda_i)^{\textstyle s_i}.$$
When $\lambda _1,\lambda_2,\lambda _3,......,\lambda _k$ are distinct eigenvalues.
We define a new matrix: $B\in M_{2n}(\mathbb R)$ by:
$$\left(\begin{matrix}
A &I_n \\ 
0 & A
\end{matrix}\right)$$
I need to prove that the minimal polynomial of $B$ is
$$M_B(t)=\prod_{i=1}^k(t-\lambda_i)^{\textstyle s_{i}+1}.$$

Edit: What follows below refered to the original version of this question, in which the definition of $B$ was:
$$B = \left(\begin{array}{cc}A&I_n\\I_n&A\end{array}\right).$$

I tried to do that in induction. I don't understand this basic case: for $1$x$1$ matrix: $\begin{pmatrix}
5
\end{pmatrix}$ we get that $ \begin{pmatrix}
5 & 1\\ 
 1& 5 
\end{pmatrix}$ 's minimal polynomial is $(x-4)(x-6)$ and it doesn't answer the condition. so I have also a problem with understanding the question I guess. 
Thanks!!

A:

For the matrix $B = \pmatrix{A & I \cr 0 & A\cr}$ and any polynomial $p$, note that $p(B) = \pmatrix{ p(A) & p'(A)\cr 0 & p(A)\cr}$.  If the polynomials $p(t)$ and $p'(t)$ 
are both divisible by $(t - \lambda)^j$, then $p(t)$ must be divisible by $(t - \lambda)^{j+1}$.

