I noticed:
$P$ is Prime ($P \geq 2$) if and only if:
$d(p^n)=d(p^m)+(n-m)$ with $n,m \in \Bbb{Z}$
where $d(x)$ is the number-of-divisors function.
The proof for it is quite simple, but I just wanted to put this out there since I couldn't find it anywhere else.
Well, yes. The divisors of $p^k$ are just the $k+1$ numbers $p^0, ..., p^k$. If you take $n>m$ you get $n-m$ new ones added onto the end of the list.