TITLE: How to compute $(1 \cdot 3 \cdot 5 \cdots 97)^2 \pmod {101}$
QUESTION [5 upvotes]: How to compute $(1 \cdot 3 \cdot 5  \cdots 97)^2 \pmod {101}$ in easiest and fastest way?

REPLY [7 votes]: I guess the only tool you currently have is the Wilson's Theorem, which says that if $p$ is a prime then $$(p-1)!\equiv-1(\mathrm{mod}~p)$$
Notice that $101$ is a prime, and $1\equiv-100(\mathrm{mod}~101),~3\equiv-98(\mathrm{mod}~101),\cdots$. Hence,
$$(1\cdot3\cdot5\cdots97)^{2}\equiv(1\cdot3\cdot5\cdots97)(1\cdot3\cdot5\cdots97)\equiv(1\cdot3\cdot5\cdots97)(-100\cdot-98\cdot-96\cdots-4)(\mathrm{mod}~101)$$
Now can you finish up?