= Wolstenholme's theorem
{c}
For every <prime number> $p\ge5$, <Wolstenholme's theorem> gives
$$
\boxed{\binom{2p-1}{p-1}\equiv1\pmod{p^3}.}
$$
Its harmonic-sum form is $H_1=\sum_{k=1}^{p-1}1/k\equiv0\pmod{p^2}$. All denominators are <coprime> to $p$, so fractions are interpreted using <modular inverses>.
Here is a proof of both assertions. Write $H_2=\sum_{k=1}^{p-1}1/k^2$. Multiplication by $2$ permutes the nonzero <residue classes> modulo $p$, so $H_2\equiv H_2/4\pmod p$. Since $p\ge5$, this implies $H_2\equiv0\pmod p$. Replacing $k$ by $p-k$ in $H_1$ gives
$$
H_1\equiv-H_1-pH_2\pmod{p^2},
$$
using $1/(p-k)\equiv-1/k-p/k^2\pmod{p^2}$. Thus $2H_1\equiv0\pmod{p^2}$, proving the harmonic-sum form.
Finally, expand the <binomial coefficient> as a product:
$$
\binom{2p-1}{p-1}=\prod_{k=1}^{p-1}\left(1+\frac pk\right)
\equiv1+pH_1+\frac{p^2}{2}(H_1^2-H_2)\pmod{p^3}.
$$
Terms containing three or more selected factors $p/k$ vanish in this <integer congruence>. The two harmonic-sum estimates make both remaining correction terms zero modulo $p^3$. For $p=5$, the result reads $\binom94=126\equiv1\pmod{125}$.
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