$NP$ is not in $P(n^k)$ for any fixed $k \geq 1$

I encountered this problem which asks to show that for any fixed $ k \geq 1$ , $ NP$ is not contained in $ P(n^k)$

As an attempt, I thought of using the time hierarchy theorem which says that there exists a language in $ P(n^{k+1})$ which is not decided in $ P(n^k)$ given that $ n^k \in o(n^{k+1})$ … Since the space of polynomial verifiers in $ NP$ is the union of all polynomials of $ n$ , using the time hierarchy theorem, this means that there exists a problem in $ NP$ that accepts instances if the accepting branch operates in time $ P(n^{k+1})$ , and so $ NP$ could not be contained in $ P(n^k)$

But is this correct ? I only assumed that such a problem exists in $ NP$ (i.e. which accepts in nondeterministic time $ P(n^{k+1})$ due to the time hierarchy theorem, but I have not really been able to construct a concrete problem…