Divisibility by $p^8$ (variation on Wolstenholme's theorem)

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.