Largest power of $p$ which divides $F_p=\binom{p^{n+1}}{p^n}-\binom{p^{n}}{p^{n-1}}$ I would like to know your comments in order to obtain the largest power of the prime number $p$ which divides
$$
F_p=\binom{p^{n+1}}{p^n}-\binom{p^{n}}{p^{n-1}}.
$$
I proved the largest power that divided $F_2$ is $3n$.
 A: For $p$ odd and $n\geq 1$,
$$
\begin{align*}
{p^{n+1}\choose p^n}-{p^n\choose p^{n-1}}&={p^n\choose p^{n-1}}\prod_{\substack{k=1\\p\nmid k}}^{p^n}\frac{p^{n+1}-k}{k}-{p^n\choose p^{n-1}}\\
&={p^n\choose p^{n-1}}\left[\prod_{\substack{k=1\\p\nmid k}}^{p^n}\left(1-\frac{p^{n+1}}{k}\right)-1\right]\\
&={p^n\choose p^{n-1}}\sum_{m\geq 1}(-1)^m p^{m(n+1)}e_m\\
&\equiv {p^n\choose p^{n-1}}\big(-p^{n+1}e_1+p^{2(n+1)}e_2\big)\mod p^{3(n+1)},
\end{align*}
$$
where $e_m$ denotes the $m$-th elementary symmetric polynomial evaluated on the set $\{k^{-1}:1\leq k\leq p^n,p\nmid k\}$. 
For $p\geq 5$ we have $e_1\equiv 0\mod p^{2n}$ and $e_2\equiv 0\mod p^{n}$ (the first statement can be found on the Wikipedia page for Wolstenholme's Theorem, the second statement is similar). Lucas' Theorem implies the largest power of $p$ dividing ${p^n\choose p^{n-1}}$ is $p$, so we conclude that $p^{3n+2}$ divides ${p^{n+1}\choose p^n}-{p^n\choose p^{n-1}}$. A larger power of $p$ is possible only if $e_1\equiv 0\mod p^{2n+1}$, which happens precisely when the numerator of the Bernoulli number $B_{p-3}$ is divisible by $p$. Only two primes with this property are known: $16843$ and $2124679$.
