For every odd number $p>1$ we have:
$\textbf{(A)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-2\qquad \textbf{(B)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p\\ \textbf{(C)}\ (p-1)^{\frac{1}{2}(p-1)} \; \text{is divisible by} \; p\qquad \textbf{(D)}\ (p-1)^{\frac{1}{2}(p-1)}+1 \; \text{is divisible by} \; p+1\\ \textbf{(E)}\ (p-1)^{\frac{1}{2}(p-1)}-1 \; \text{is divisible by} \; p-1$
Given that $p$ is odd, $p-1$ must be even, therefore ${{\frac{1}{2}}(p-1)}$ must be an integer, which will be denoted as n.
\[(p-1)^n-1\]
By sum and difference of powers
\[=((p-1)-1)((p-1)^{n-1}+\cdots+1^{n-1})\]
\[=(p-2)((p-1)^{n-1}+\cdots+1^{n-1})\]
$p-2$ divide $(p-1)^{{\frac{1}{2}}(p-1)}$$\fbox{A}$.