Originally Posted by
melese But, for example, $\displaystyle 3\cdot4\equiv1\pmod{11}$, and $\displaystyle \displaystyle3,4\leq\frac{11-1}{2}$ (If I understand what you meant.).
I don't think you do, but for sure I don't understand you: "inverse" means "additive inverse", not "multiplicative inverse", as
I think it was clear from the context of my answer, so your example is irrelevant, I believe, and what is relevant is what you wrote
after that, which is what I was trying the OP to realize by himself.
Tonio
The observation here is that for $\displaystyle \displaystyle1\leq x\leq\frac{p-1}{2}$, we have $\displaystyle \displaystyle\frac{p+1}{2}\leq p-x\leq p-1$.
Then noting that $\displaystyle p-x\equiv-x\pmod p$ shows that $\displaystyle \displaystyle\frac{p+1}{2}\cdots(p-2)(p-1)\equiv(-\frac{p-1}{2})\cdots(-2)(-1)\pmod p$.