I would appreciate some insight. The difficult part for me is proving that the homomorphism $\Phi$ is surjective.
Consider the group action of $GL_k(\mathbb{Z}_p)$ acting on $\mathbb{(Z_p)^k}$ by matrix multiplication, which induces the homomorphism $\Phi: GL_k(\mathbb{Z_p})\to S_{p^k}$ (since ord$[(\mathbb{Z}_p)^k]=p^k$). This action is faithful and thus $\Phi$ is injective.
Now the final step is to prove that $\Phi$ is surjective, which can be done by arguing that ord$[GL_k(\mathbb{Z_p})]=p^k!$. Or is there a better way?
Resolved: $GL_k(\mathbb{Z}_p)\cong H$, where $H < S_{p^k}$ by virtue of the First Isomorphism Theorem: $GL_k(\mathbb{Z}_p)/\{e\} \cong im(\Phi)$.
It's not surjective. If it were surjective, that would mean that every bijection $\mathbb{Z}_p^k\to\mathbb{Z}_p^k$ is a linear map, which is certainly false (as long as $k>0$). This shows that $|GL_k(\mathbb{Z}_p)|<p^k!$, so $GL_k(\mathbb{Z}_p)$ cannot be isomorphic to $S_{p^k}$ by any other map either.
However, it is of course surjective when considered as a map to its image, so it gives an isomorphism from $GL_k(\mathbb{Z}_p)$ to the subgroup $\operatorname{im}(\Phi)$ of $S_{p^k}$.