The eigenvalues of $A^k$ are $\lambda_1^k, \lambda_2^k, \ldots, \lambda_n^k$

$\text{tr}(A^k) = \lambda_1^k + \lambda_2^k + \cdots + \lambda_n^k$

$\det(A^k) = (\det A)^k = (\lambda_1 \lambda_2 \cdots \lambda_n)^k$

If $A$ has eigenvalue 0, then $A^k$ is invertible for all $k \geq 1$