Concerning a minimal polynomial I'm currently taking linear algebra and I'm working on a questing concerning minimal polynomial.
The question is as follows:

Consider a matrix $A$, whose minimal polynomial is of the form $p(x)^n$ where $p$ is irreducible and $n \geq 1$ is an integer.
We are to show that there exists a vector $\vec v$ such that its $A$-minimal polynomial is precisely $p(x)^n$.

The definition given for a $A$-minimal polynomial is a polynomial $m$ with minimum degree such that $m(A)\vec v=0$.
The question also has a second part which is:

There exists a vector $\vec v$ whose $A$-minimal polynomial is the matrix $A$'s minimal polynomial.

A hint is given which tells us to use the Primary Decomposition Theorem.
My Attempt
I recall that the definition of minimal polynomial is the factor with least degree of the characteristic polynomial which still nullifies $A$. It follows that $p(A)^n=0$ but $p(A)^k\neq 0,\ k<n$.
I can see that for such $k$, $p(A)^kp(A)^{n-k}=0$ but that doesn't lead me to the equation I'm looking for.
For the second part I recall the Spectral Theorem which implies that any matrix in the complex numbers is diagonalizable and the corresponding vector space has a basis consisting of eigenvectors.
I'm a bit rusty with the other theorem and I'm not sure how to use it along with the result of the first part in order to prove the second part.
I'm looking for hints, comments, or partial solutions since I'm currently stuck.
Any help will be greatly appreciated.
 A: For the first part, it serves to know that what you call the $A$-minimal polynomial of a vector$~v$ (i.e., the minimal degree monic polynomial$~P$ such that $v\in\ker(P[A])$) is always a divisor of the (full) minimal polynomial of $A$; this is true because all polynomials $Q$ for which $v\in\ker(Q[A])$ are polynomial multiples of $P$, by the usual Euclidean division argument: the remainder of dividing $Q$ by $P$ has the same property, but degree less than $\deg(P)$, so it must be zero. So for all vectors this $A$-minimal polynomial is some $p^k$ with $k\leq n$; can you see why one cannot have $k<n$ for all vectors?
For the second part, use the (canonical) decomposition of the vector space into the kernels of the primary divisors of the minimal polynomial, evaluated at$~A$; in each of the subspaces the minimal polynomial of the restriction of (the linear operator defined by)$~A$ is that primary divisor, so the first part applies there. Then combine vectors in the subspaces obtained from the first part into a vector that works for all of the minimal polynomial.
