Proof of statement about invertible matrices raised to the power k

Question

True or false?

Let $A$ be a matrix of size $n \times n$. If there is a natural number $k$ such that $A^k$ is invertible, then $A$ is invertible.

I intuitively understand that this is correct, but I would like to know what the formal proof for this statement is…

Thank you!

Answer 1

Hint

If the eigenvalues of the matrix $A$ are $\lambda_1,\cdots ,\lambda_n$, then the eigenvalues of matrix $A^k$ would be $\lambda_1^k,\cdots ,\lambda_n^k$

Answer 2

If $A^k$ is invertible then $\det (A^k)\ne 0$. But $\det(A^k)=\det(A)^k$ (since a field has no zero divisors, so $\det(A)^k\ne 0$ or $\det(A)\ne 0$, meaning $A$ is invertible.