For my purposes, you may want to interpret "best" as "clearest and easiest to understand for undergrads in a first number theory course," but don't feel too constrained.
I think by far the simplest easiest to remember elementary proof of QR is due to Rousseau (On the quadratic reciprocity law).  All it uses is the Chinese remainder theorem and Euler's formula $a^{(p-1)/2}\equiv (\frac{a}{p}) \mod p$.  The mathscinet review does a very good job of outlining the proof.  I'll try to explain how I remember it here (but the lack of formatting is really rough for this argument).
Here's the outline.  Consider $(\mathbb{Z}/p)^\times \times (\mathbb{Z}/q)^\times = (\mathbb{Z}/pq)^\times$.  We want to split that group in "half", that is consider a subset such that exactly one of $x$ and $-x$ is in it.  There are three obvious ways to do that.  For each of these we take the product of all the elements in that "half."  The resulting three numbers are equal up to an overall sign.  Calculating that sign on the $(\mathbb{Z}/p)^\times$ part and the $(\mathbb{Z}/q)^\times$ part give you the two sides of QR.
In more detail.  First let me describe the three "obvious" halves:

Take the first half of $(\mathbb{Z}/p)^\times$ and all of the other factor.
Take all of the first factor and the first half of $(\mathbb{Z}/q)^\times$.
Take the first half of $(\mathbb{Z}/pq)^\times$.

The three products are then (letting $P = (p-1)/2$ and $Q=(q-1)/2$):

$(P!^{q-1}, (q-1)!^P)$.
$((p-1)!^Q, Q!^{p-1})$.
$\left(\dfrac{(p-1)!^Q P!}{q^P P!},\dfrac{(q-1)!^P Q!}{p^Q Q!}\right)$.

All of these are equal to each other up to overall signs.  Looking at the second component it's clear that the sign relating 1 and 3 is $\left(\frac{p}{q}\right)$.  Similarly, the sign relating 2 and 3 is $\left(\frac{q}{p}\right)$.  So the sign relating 1 and 2 is $\left(\frac{p}{q}\right) \left(\frac{q}{p}\right)$.  But to get from 1 to 2 we just changed the signs of $\frac{p-1}{2}  \frac{q-1}{2}$ elements.  QED
