If $p>3$ a prime number then $\binom {p-1}{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} 4^{p-1} \pmod {p^3}$

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Here is one of Morley's theorem in number theory.

My idea is to begin in $\mathbb{Z}/p\mathbb{Z}$ :

$\binom {p-1}{\frac{p-1}{2}} = \frac{(p-1)!}{(\frac{p-1}{2})!(\frac{p-1}{2})!}=\frac{(p-1)!}{(-1)^{\frac{p-1}{2}}(p-1)!}=\frac{1}{(-1)^{\frac{p-1}{2}}}\equiv (-1)^{-\frac{p-1}{2}} \pmod p.$

Now, I want to build a solution for $p^2$ then for $p^3$ with maybe Hensel's lemma but I don't know if it's the right way or if it can work.

Thanks in advance !

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Just a partial answer, for now. By Wolstenholme's theorem we have $\binom{2p-1}{p-1}\equiv 1\pmod{p^3}$, and

$$\begin{eqnarray*} \binom{2p-1}{p-1} &=& \frac{\color{red}{(2p-1)\cdot (2p-3)\cdot\ldots\cdot(p+2)}}{\color{blue}{(p-2)\cdot(p-4)\cdot\ldots\cdot 1}}\cdot\binom{p-1}{\frac{p-1}{2}}\\ &=&(-1)^{\frac{p-1}{2}}\binom{p-1}{\frac{p-1}{2}}\prod_{k=1}^{\frac{p-1}{2}}\left(1-\frac{2p}{2k-1}\right)\tag{1}\end{eqnarray*} $$ but $q(z)=\prod_{k=1}^{\frac{p-1}{2}}\left(1-\frac{2z}{2k-1}\right)$ is a polynomial that fulfills $q(0)=1$ with roots at $z\in\left\{\frac{1}{2},\frac{3}{2},\ldots,\frac{p-2}{2}\right\}$. By Viète's theorem, the coefficient of $z$ in $p(z)$ just depends on: $$ \frac{2}{1}+\frac{2}{3}+\ldots+\frac{2}{p-2} = 2H_{p-1}-H_{\frac{p-1}{2}}\tag{2}$$ while the coefficient of $z^2$ just depends on: $$ \frac{1}{2}\left[\left(2H_{p-1}-H_{\frac{p-1}{2}}\right)^2-\left(4H_{p-1}^{(2)}-H_{\frac{p-1}{2}}^{(2)}\right)\right] \tag{3}$$ so Morley's theorem ultimately boils down to understanding the behaviour of $H_{\frac{p-1}{2}}\pmod{p^2}$ and $H_{\frac{p-1}{2}}^{(2)}\pmod{p}$, just like Wolstenholme's theorem is equivalent to $H_{p-1}\equiv 0\pmod{p^2}$ and $H_{p-1}^{(2)}\equiv 0\pmod{p}$.

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In the literature, Morley's result has been published in Annals of Math 9, 1895. He showed the congruence using an explicit form of de Moivre's Theorem. For another proof see the paper Binomial coefficients modulo prime powers by Andrew Granville, section $9$ and the remarks on equation $(14)$. Here perhaps a better reference for the proof is Theorem $6$ in the paper On the Residues of Binomial Coefficients and Their Products Modulo Prime Powers by Cai and Granville.
A further proof is given in the arXiv-paper Morley's other miracle by Aebi and Cairns, which is freely available.