Let $n$ be a natural and $K$ a field. When can we find an $n\times n$ invertible matrix $A$ of order $n+1$? Order here means $A^{n+1}=I$ and is the smallest integer that satisfies it.
This is what I have achieved:
Matrix $A$ will fulfill $A^{n+1}-I=0$. This implies the minimal polynomial of $A$ will divide $x^{n+1}-1=(x-1)(x^n+\cdots+1)$. If $n+1$ is prime, this is all we can factor $x^n-1$. So necessarily $p=x^n+\cdots+1$ and $A$ will be similar to the matrix associated with this polynomial.
What else can we say?
The part
from your question is wrong, $GK_n(p^k)$ is never a $p$-group if $k>0, n>0$.
If $K$ has characteristic $0$, your matrix $A$ exists for every $n$. Note that $K$ contains a copy of $\Bbb Q$. So we assume $\Bbb Q\le K$.
Case 1. $n+1$ is not a prime, $n+1=km$, $k,m>1$. Let $\alpha$ and $\beta$ be roots of $1$ equal to $\exp(2\pi i/k)$ and $\exp(2\pi i/m)$ respectively. Let $p_k$ and $p_m$ be the minimal polynomials for these numbers over $\Bbb Q$. Let $A$ be the companion matrix of $p_kp_m$ over $\Bbb Z$ whose eigenvalues are the roots of $p_kp_m$. Then the order of that matrix is $km$. To make $A$ an $n\times n$-matrix, add some diagonal $1$'s: $\mathrm{diag}(A,1,...,1)$.
Case 2. $n+1$ is a prime. Then $(x^{n+1}-1)/(x-1)=p(x)$ is irreducible of degree $n$ over $\Bbb Q$, so it has $n$ different complex roots, take the companion matrix $A$ over $\Bbb Z$ with minimal polynomial $p$, its order is $n+1$.