Let s be half of p-1, and let the equations run from 1 to s.  Multiply the s equations together, hence the left side becomes s!.  The right side includes a 2 from the second equation, a 4 from the fourth equation, and so on.  Once we pass s, the right side still has the even numbers, but they are disguised as negative numbers.  Twice s shows up as -1 (first equation), twice (s-1) is -3 (third equation), and so on.  So the right side contains s!×2^s.  The s! cancels the s! on the left, and 2^s is the same as [2\p].  Finally, the right side includes (-1)^1+2+…+s.  The exponent simplifies to s×(s+1)/2.  This is even when s is 3 or 4 mod 4, which means 2 is a quadratic residue iff p = 1 or 7 mod 8.  2 is the square of 6 mod 17, but it isn't the square of anything mod 13.