Next consider [2\p]. Write the following series of equations.
1 = (-1)×(-1)1
2 = 2×(-1)2
3 = (-3)×(-1)3
4 = 4×(-1)4
…
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!×2s. The s! cancels the s! on the left, and 2s 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.
If 2 is a square mod p then it is a square in F. Since p is 1 or 7 mod 8, the same is true of w, and we're in good shape.
If 2 is not a square, and p is 3 or 5 mod 8, let F have odd dimension over p, whence w is 3 or 5 mod 8. Adjoin the square root of 2 and find an intermediate extension of dimension 2. This has to be a factor of the dimension of F over p, and that is impossible. Hence 2 is not a square.
Finally let F have even dimension over p. Thus w is 1 or 7 mod 8. Since the extension is galois, it includes the extension of dimension 2. Restrict F to this extension. The cyclic multiplicative group has order p2-1. The discrete log of each integer is a multiple of p+1. This is even, hence every integer is a square, including 2.