<s>[INST] If $p>3$ a prime number then $\binom {p-1}{\frac{p-1}{2}} \equiv (-1)^{\frac{p-1}{2}} 4^{p-1} \pmod {p^3}$ 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 !
 [/INST] 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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