Given $A$ is invertible, how to derive the characteristic and minimal polynomial of $A^n$? Given A an nxn invertible complex matrix; show how to derive the characteristic and minimal polynomials of $A^n$ from the characteristic and minimal polynomials of $A$. 
I can see how easy this would be if $A$ were diagonalizable, but I don't understand what useful information '$A$ is invertible' gives us?  
 A: As akech wrote, the roots of $q(y)=\det(A^k-yI_n)$ are the $(\lambda_i^k)_i$ where the $(\lambda_i)_i$ are the roots of $p(x)=\det(A-xI_n)$; in particular, the previous polynomials have same degree. Thus it suffices to eliminate $x$ in the system $p(x)=0,x^k-y=0$. We can do that with hand or using the Maple library "Grobner basis" (recall that this software works over a commutative ring). For instance, when $p(x)=x^3-5x^2+4x-3$ and $k=13$, we obtain $q(y)=y^3-134892776y^2-1094609y-1594323$.
The case of the minimal polynomial $m_A$ of $A$ is more complicated because the degree may vary. We must see if there are $\lambda_i\not=\lambda_j$ s.t. $\lambda_i^k=\lambda_j^k$. For instance, let $A=diag(I_2+J_2,-I_3+J_3)$ where $J_p$ is the nilpotent Jordan block of dimension $p$; its minimal polynomial is $(x-1)^2(x+1)^3$; on the other hand, $A^2$ has $(y-1)^3$ as minimal polynomial.
Anyway, the first step is to eliminate $x$ in the system $m_A(x)=0,x^k-y=0$. Then $m_{A^k}(y)$ is a divisor of the result; according some tests with Maple, it seems that the result of elimination is the required minimal polynomial; to be checked... 
It seems also that the hypothesis "$A$ invertible" is useless; for instance, the elimination of $x$ in $\{x^{50}=0,x^{13}-y=0\}$ gives $y^4=0$.
