Several people above have mentioned the proof above using Gauss sums. In a recent assignment for our Galois theory course, we were asked to prove quadratic reciprocity using Galois theory. I would like to remark that in determining the fixed field of degree 2 over $\Bbb{Q}$ corresponding to the subgroup of index $2$ in $\textrm{Gal}(\Bbb{Q}(\zeta_p)/\Bbb{Q})$, one does not need to know about Gauss sums. Instead notice that $\Bbb{Q}(\zeta_p)$ is the splitting field of $x^p -1$, so that the discriminant of this polynomial $(-1)^{p(p-1)/2}p^p$ is in $ \Bbb{Q}(\zeta_p)$. The discriminant can be computed easily using the Vandermonde determinant and some knowledge about power sums.
Since the square root of the discriminant is a product of differences of roots, it too is in the splitting field. Hence we see that $\Bbb{Q}(\sqrt{\pm p })$ is the unique subfield of index 2 contained in $\Bbb{Q}(\zeta_p)$, depending on whether $p \equiv 1$ or $3$ mod $4$.
Keith Conrad remarked above that in the usual proof using Gauss sums, we need to show that $1 \equiv -1 \mod(p\Bbb{Z}[\zeta_p])$ gives us our desired contradiction, or that $2/p \in \Bbb{Z}[\zeta_p]$ gives a contradiction. I believe one can do this without invoking anything about integral bases. Instead we can say by Proposition 5.1(iii) of Atiyah - Macdonald that $2/p$ is integral over $\Bbb{Z}$ and since $\Bbb{Z}$ is integrally closed in its field of fractions (the proof of which can be rephrased entirely without any commutative algebra) this forces $2/p$ to be an integer. But by assumption $p$ was a prime not equal to 2 so this is a contradiction.
