Show that $|G|=1000$ is not simple

Question

I think I've done this, but I just want to check it's proved enough.

So $|G|=1000=8\cdot125$.

Then $n_5\in\{\text{factors of 8}\}$ so $n_5\in\{1,2,4,8\}$ congruent to $1\bmod5$.

$2$, $4$, $8$ are not congruent to $1\bmod5$.

So $n_5=1$ and if $n_p=1$ then the Sylow $p$-subgroup is unique and normal so $G$ cannot be simple.