We prove that if a tile in $\mathbb Z^d$ has prime size $p$, then it must be spectral. The proof is by contradiction, it is simply shown that the tiling complement of such a tile can not annihilate all $p$-subgroups. In addition, with a simple transformation we prove that any $p$ points in general linear positions in $\mathbb Z^d (d\ge p-1)$ must be both tiling and spectral.