Is there a proof of quadratic reciprocity using $p$-adic numbers? I asked same question on MSE before, but I didn't get any answer yet. 

I know that the quadratic reciprocity can be regarded as a special case of Artin reciprocity (class field theory), and we can get it by considering the cyclotomic extension of $\mathbb{Q}_{p}$. However, I want to know if there's any proof of quadratic reciprocity that doesn't use any stronger result, but only uses some properties of $p$-adic numbers. Thanks in advance. 
More precisely, we can do analysis on $\mathbb{Q}_{p}$. We have an exponential and logarithm function on $\mathbb{Q}_{p}$ (at least for $p>2$), and we understand unit group $\mathbb{Z}_{p}^{\times}$ well, etc. 
But I don't know how to prove it in a purely local and analytic way.
 A: There are multiple proofs of quadratic reciprocity via finite fields, including Gauss's third proof. One of my favorites is Zolotareff's proof based on his interpretation of Legendre symbol as the sign of a permutation ("Zolotarev's lemma"). This approach seems to fit your criteria, because $\Bbb{F}_p\simeq\Bbb{Z}_p/p\Bbb{Z}_p$ as rings ("we understand unit group $\Bbb{Z}^{\times}_p$ well") and one may inflate functions from $\Bbb{F}_p$ to $\Bbb{Z}_p$.
A: The quadratic reciprocity law is a special case of the product formula for Hilbert's symbol: for all $(a,b)\in\mathbb{Q}^{\times}$
\begin{equation}
\underset{v}{\prod}(a,b)_{v}=1
\end{equation}
where $v$ ranges in the product other all prime numbers and $\infty$. The definition of the symbol $(a,b)_{p}$ can be given in purely $p$-adic terms, but of the course the proof of the product formula must involve global arguments, so is global in nature (as any proof of the quadratic reciprocity law must be at least in the sense that the statement of this theorem involves two primes). When done through the structure of the Brauer group and cyclic algebras, the proof of the product formula is however minimally global in some sense.
Is that a satisfying answer for you?
