Eisenstein and Quadratic Reciprocity as a consequence of Artin Reciprocity, and Composition of Reciprocity Laws Question 1: I've heard that Eisenstein and Quadratic Reciprocity can be derived from the Artin Reciprocity by applying it to certain field extensions. But I haven't seen on any reference an explicit description of this, and I am here asking for one.
Question 2: I have seen quartic, octic, and sextic reciprocity laws. They look just like by applying some kind of power reciprocity in fields satisfying another power reciprocity law. So is there a general procedure to construct reciprocity of arbitrary powers based on composition of reciprocity of degree 2 and odd primes (Since their product generate the natural numbers)? If not, what is the obstruction?
 A: To derive quadratic reciprocity from Artin reciprocity, consider the field extensions $$\mathbb{Q} \subset F=\mathbb{Q}(\sqrt{p^*}) \subset K=\mathbb{Q}(\zeta)$$ where $p$ is a prime $\ne 2$, $\zeta$ is a primitive $p$th root of unity, and $p^* = (-1)^{(p-1)/2} p$.  
(To see that $F$ is contained in $K$, look up Gauss sums.)
Let $q$ be another prime, $q \ne p, q \ne 2$.  We know that $p^*$ is a square mod $q$ iff $q$ splits in $F$ iff the Artin symbol of $q$ in $F/\mathbb{Q}$ is trivial.  (All Galois groups considered here are abelian, so the Artin map depends only on the base prime $q$ and not on any particular prime in the upper field.) 
The Artin symbol of $K/\mathbb{Q}$ over the prime $q$ is the element $\sigma_q: \zeta \mapsto \zeta^q$ in the Galois group of $K/\mathbb{Q}$.  
But Artin($F/\mathbb{Q}$) is the restriction of Artin($K/\mathbb{Q}$) to $F$, so Artin($F/\mathbb{Q}$) is trivial iff $\sigma_q$ is in the kernel of this restriction map.  Both Galois groups are cyclic, and so this restriction map is the unique nontrivial homomorphism from $(\mathbb{Z}/p\mathbb{Z})^{\times}$ to $\{\pm 1\}$.  It is easy to check that $\sigma_q$ will be in the kernel of the restriction map iff $q$ is a square mod $p$.
Conclusion: $p^*$ is a square mod $q$ iff $q$ is a square mod $p$.  This is quadratic reciprocity.
For other reciprocity laws, you'll have to choose the field $F$ differently.  For example, for cubic reciprocity, choose $F$ to be the unique degree 3 extension in $K/\mathbb{Q}$.  (Of course, this only exists if $p \equiv 1$ (mod 3), but if that condition fails then cubic reciprocity is not very interesting: everything will be a cube mod $p$.)  Things get more complicated because the analog of the step "$p^*$ is a square mod $q$ iff $q$ splits in $F$" is not as simple.  But it's the same idea.
Edit: This last paragraph is not quite correct.  The proper extensions to consider for cubic reciprocity is $L \subset L(p^{1/3})$ where $L = \mathbb{Q}(\omega)$, $\omega$ a primitive cube root of unity.  The Artin conductor of this extension is a divisor of $3p$, and given a prime $q \equiv 1$ (mod 3) in $L$ (relatively prime to $3p$), the image of the ideal $(q)$ given by Artin reciprocity is essentially the unique nontrivial map from the cyclic group $(O_L / q)^{\times}$ to the cyclic group Gal($L(p^{1/3})/L$) of order 3.  For details, see the senior thesis of Noah Snyder referenced by KCd in the comments above.  For general $n$th power reciprocity, use $L \subset L(p^{1/n})$ where $L$ is the cyclotomic field generated by a primitive $n$th root of unity.
A: For a nice simple example use the extension $\mathbb{Q}(i)/\mathbb{Q}$. 
Notice that for an odd prime $p$ (i.e. unramified) the Frobenius element of $p$ relative to this extension is the identity if and only if $p\equiv 1 \bmod 4$.
When showing this algebraically you find that:
Frob$_p(a+ib) \equiv a + (-1)^{\frac{p-1}{2}}bi \bmod \mathfrak{p}$
However, by Eulers criterion the RHS is congruent to $a + \left(\frac{-1}{p}\right)bi\bmod \mathfrak{p}$. 
So really the Frobenius element of $p$ here encodes the value of the Legendre symbol $\left(\frac{-1}{p}\right)$.
Thus $\left(\frac{-1}{p}\right) = 1$ if and only if Frob$_p$ is the identity, which is equivalent to $p \equiv 1 \bmod 4$.
This gives one of the supplementary laws. The rest of quadratic reciprocity comes from changing the above extension to something similar (as shown by the other replies).
