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For every prime number $p\ge 5$, Wolstenholme's theorem givesIts harmonic-sum form is $H_1={\sum }_{k=1}^{p-1}1/k\equiv 0(\mathrm{mod}{p}^2)$. 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 $H_2\equiv H_2/4(\mathrm{mod}p)$. Since $p\ge 5$, this implies $H_2\equiv 0(\mathrm{mod}p)$. Replacing $k$ by $p-k$ in $H_1$ givesusing $1/(p-k)\equiv -1/k-p/k^2(\mathrm{mod}{p}^2)$. Thus $2H_1\equiv 0(\mathrm{mod}{p}^2)$, proving the harmonic-sum form.

Finally, expand the binomial coefficient as a product: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 $\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126\equiv 1(\mathrm{mod}125)$.