Number of conjugacy classes in group of prime power order is congruent to order of group modulo prime-square minus one

Statement
Suppose $$p$$ is a prime number and $$P$$ is a group of prime power order with underlying prime $$p$$. Then the number of conjugacy classes of $$P$$ (which is the same as the number of irreducible representations) is congruent to the order of $$P$$ modulo $$(p^2 - 1)$$.

Facts used

 * 1) uses::Number of irreducible representations equals number of conjugacy classes
 * 2) uses::Degree of irreducible representation divides order of group: For a $$p$$-group, in particular, this means that it is a power of $$p$$.
 * 3) uses::Sum of squares of degrees of irreducible representations equals order of group

Proof
Given: A prime number $$p$$. A finite $$p$$-group $$P$$ of order $$p^r$$ with $$n$$ conjugacy classes.

To prove: $$n \equiv p^r \pmod{(p^2 - 1)}$$

Proof: