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

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$n$ equals the number of irreducible representations of $P$.	Fact (1)	$n$ is the number of conjugacy classes.
2	All irreducible representations of $P$ have degree $p^i$ for some nonnegative integer $i$, with $0 \le i \le r$.	Fact (2)	$P$ has order $p^r$		[SHOW MORE] All the degrees divide $p^r$, and any divisor of $p^r$ must be a power $p^i$ with $0 \le i \le r$. (In fact, we can do better and get that $i < r/2$ using Fact (3), but we do not need that bound for the present purpose).
3	Let $a_i$ be the number of irreducible representations of degree $p^i$ for $i = 0,1,2,\dots,r$. Then, $a_0 + a_1 + a_2 + \dots + a_r = n$.			Steps (1), (2)	Step-combination direct.
4	$a_0(p^0)^2 + a_1(p^1)^2 + a_2(p^2)^2 + \dots + a_r(p^r)^2 = p^r$	Fact (3)	$P$ has order $p^r$	Step (3)	Step-fact direct.
5	$a_i \equiv a_i(p^i)^2 \pmod{(p^2 - 1)}$				[SHOW MORE] We know that $1 \equiv p^2 \pmod{p^2 - 1}$. Raise both sides to the power of $i$ and multiply by $a_i$.
6	$a_0 + a_1 + a_2 + \dots + a_r \equiv a_0(p^0)^2 + a_1(p^1)^2 + a_2(p^2)^2 + \dots + a_r(p^r)^2 \pmod{(p^2 - 1)}$			Step (5)	Sum up Step (5) for $0 \le i \le r$.
7	$n \equiv p^r \pmod{(p^2 - 1)}$			Steps (3), (4), (6)	Step-combination direct.