Divisibility by $p^8$ (variation on Wolstenholme's theorem) Prove that for primes $p≥5$: $\binom{p^3}{p^2}-\binom{p^2}{p} ⋮ p^8$. Three vertical dots stand for "is divisible by".
Thoughts:
Let $p≥5$ be a prime. Then for $n≥2$, the $p$-valuation of $\binom{p^n}{p^{n-1}}-\binom{p^{n-1}}{p^{n-2}}$ is $3n−1$. Here the $p$-valuation of an integer $m$ is $\nu_{p}(m)=\max \{k: p^{k} \mid m\}$. In fact, I can't even see a simple proof for $n=2$, as it seems that there are at least $2$ powers of $p$ that are not "obvious".
I suspect that a proof is somehow obtainable using a refinement of the following well known observation that is true in any commutative ring: If $x \equiv y \operatorname{mod} p^{k}$, then $x^{p} \equiv y^{p} \operatorname{mod} p^{k+1}$. The idea is to use it somehow, starting with $(1+x)^{p} \equiv 1+x^{p} \operatorname{mod} p$. But I don’t see how to make this work. Any ideas?
 A: I'll prove the general version. Let $S$ be the set of integers at most $p^{n-1}$ that are relatively prime to $p$. We compute
$$\binom{p^n}{p^{n-1}}=\prod_{i=1}^{p^{n-1}}\frac{p^n-i}{i}=\prod_{i\in S}\frac{p^n-i}{i}\prod_{i=1}^{p^{n-2}}\frac{p^{n-1}-i}{i}=\binom{p^{n-1}}{p^{n-2}}\prod_{i\in S}\left(1-\frac{p^n}i\right),$$
where we have used that $|S|=p^{n-2}(p-1)$ is even. Since
$$\nu_p\left(\binom{p^{n-1}}{p^{n-2}}\right)=1,$$
we need to show that
$$\prod_{i\in S}\left(1-\frac{p^n}i\right)\equiv 1\pmod{p^{3n-2}}.$$
We expand
$$\prod_{i\in S}\left(1-\frac{p^n}i\right)=1-p^n\sum_{i\in S}\frac1i+p^{2n}\sum_{\{i_1,i_2\}\subset S}\frac1{i_1i_2}\pmod{p^{3n}}\tag{1}.$$
Let
$$s_1=\sum_{i\in S}\frac1i\text{ and }s_2=\sum_{i\in S}\frac1{i^2}.$$
We first claim that $s_1\equiv s_2\equiv 0\pmod{p^{n-1}}$. Indeed, since $S$ is closed under inversion (modulo $p^{n-1}$),
$$s_1\equiv \sum_{i\in S}i\pmod{p^{n-1}}\text{ and }s_2\equiv \sum_{i\in S}i^2\pmod{p^{n-1}}.$$
We can explicitly calculate, if $q=p^{n-1}$,
$$s_1=\frac{q(q+1)}2-p\frac{q/p(q/p+1)}2\equiv 0\pmod q$$
and
$$s_2=\frac{q(q+1)(2q+1)}{6}-p^2\frac{(q/p)(q/p+1)(2q/p+1)}{6}\equiv 0\pmod q,$$
using that $p\not\in\{2,3\}$. Now, we have by (1) that
$$\prod_{i\in S}\left(1-\frac{p^n}i\right)\equiv 1-p^ns_1+\frac{p^{2n}(s_1^2-s_2)}{2}\equiv 1-p^ns_1\pmod{p^{3n-1}}.$$
So, we need only to show that
$$s_1\equiv 0\pmod{p^{2n-2}}.$$
Now,
$$2s_1=\sum_{i\in S}\frac1i+\sum_{i\in S}\frac1{p^{n-1}-i}=p^{n-1}\sum_{i\in S}\frac1{i(p^{n-1}-i)}\equiv -p^{n-1}s_2\pmod{p^{2n-2}}.$$
Since $s_2\equiv 0\pmod{p^{n-1}}$, this is $0$, so $s_1\equiv 0\pmod{p^{2n-2}}$, as desired.
