diagonalizability of matrix If $A$ is invertible, $F$ is algebraically closed, and $A^n$ is diagonalizable for some $n$ that is not an integer multiple of the characteristic of $F$, then $A$ is diagonalizable.
My question is: 
(1) Why the condition "$A$ is invertible" essential?
(2) In wikipedia, diagonalizable matrix , it says $A$ satisfies the polynomial $p(x)=(x^n- \lambda_1) \cdots (x^n- \lambda_k)$, is that means $p(x)$ is the minimal polynomail of $A^n$? If yes, why it can be written in the form of $(x^n- \lambda_1) \cdots (x^n- \lambda_k)$ instead of $(x- \lambda_1) \cdots (x- \lambda_k)$?? 
Thank you very much!

If $A$ is invertible then $A^n$ is also invertible, so $0$ is not an eigenvalue of $A^n$. 
$A^n$ is diagonalizable then  its minimal polynomial $P$ is a product of distinct linear factors over $F$:
$$P(X)=(X-\lambda_1)\cdots(X-\lambda_k),\quad \lambda_i\neq\lambda_j$$
We know that  $A^n$ is annihilated by $P$: $P(A^n)=0$ so $A$ is annihilated by the polynomial:
$$Q(X)=(X^n-\lambda_1)\cdots(X^n-\lambda_k)$$
and since $\lambda_i\neq0$ then each polynomial $(X^n-\lambda_i)$  has no multiple root (since its derivative is $nX^{n-1}$ and $n$ is not an integer multiple of the characteristic of $F$), then $Q$ is a product of distinct linear factors and $A$ is annihilated by $Q$. Hence $A$ is diagonalizable matrix.
