Is there a natural number $k$ such that for every prime $p$ there is a non-Abelian group of order $p^k$?

Question

Is there a natural number $k$ such that for every prime $p$ there is a non-Abelian group of order $p^k$?

Update: $k=3$ should work; there is a nontrivial semidirect product $(\mathbb{Z}/p^2\mathbb{Z}) \rtimes (\mathbb{Z}/p\mathbb{Z})$, which follows from $\mathrm{Aut}(\mathbb{Z}/p^2\mathbb{Z})$ having an order $p$ element, since that automorphism group has order $\phi(p) = p(p-1)$ which is divisible by $p$. (Thanks to those who helped point out my mistake in computing the order of $\mathrm{Aut}$ of a cyclic group.)

Answer 1

Hint :

$UT(n, \Bbb{R})$ denote set of all upper triangular matrices with all entries on the diagonal are $1$.

Then $|UT(n, \mathbb{F}_p) |=p^n$