Wilson's Theorem Factorial

Question

I need to prove that $ (1 \cdot 3 \cdot 5 \dotsm 2009)^2 - 1 \equiv 0 \pmod{2011}$

By modular simplification, I need to prove that $(3 \cdot 5 \cdot 7 \dotsm 2009) \equiv 1 \pmod{2011}$

I know that I need to link this to Wilson's theorem, so can someone nudge me in the right direction in the comments, and provide a solution?

Answer 1

Since $2011$ is prime, Wilson's Theorem applies. Modulo $2011$ we have $$\eqalign{ (1\times3\times\cdots\times2009)^2 &=1\times3\times\cdots\times2009\times 1\times3\times\cdots\times2009\cr &\equiv1\times3\times\cdots\times2009 \times(-2010)\times(-2008)\times\cdots\times(-2)\cr &\equiv(-1)^{1005}(2010)!\cr &\equiv1\ .\cr}$$