Torque, and the Law of the Lever How fundamental is the Law of the Lever? It seems that we simply define torque as being $r \times F$, if that's the case, then torque isn't a derived quantity, is it? Something like the Law of the Lever is distinct from Newton's Laws, then?
 A: General remarks.
That's right.
Torque is defined as $\mathbf r\times\mathbf F$ where $\mathbf r$ is the position where the force is applied, and $\mathbf F$ is the force being applied.
The so-called Law of the Lever can then be derived from the following fact (which itself can be derived from Newton's Laws) about systems of particles:

The net external torque on a system of particles equals the rate of change of its total angular momentum;
  \begin{align}
  \boldsymbol\tau_\mathrm{ext} = \frac{d\mathbf L}{dt}.
\end{align}

For a proof, see
https://physics.stackexchange.com/a/55255/19976
Strictly speaking, as noted in the linked post, for this to be true, the particles in the system must interact according to forces that point in the direction of the lines joining them, but for macroscopic objects like levers, this condition is fulfilled.
Emergence of the Law of the Lever.
To see how this gives the Law of the lever, consider a light but very strong and rigid lever that is not rotating at all with a fulcrum that is a distance $d$ from one end, and a larger distance $D$ from the other.  Suppose that you were to hang a mass $m$ from the end that is closer to the fulcrum.  
If the lever is horizontal, what force $F$, would you need to exert on the opposite end so as to make it balance and therefore hold up the mass on the other side?  
Well if we take the location where the fulcrum makes contact with the lever to be the origin, then the torque due to the hanging mass is $mgd$ while the force you exert on the other end would have a corresponding torque $-FD$.  They have opposite sign because they tend to rotate the lever in opposite directions about the fulcrum.  Since the level will be standing still, it's angular momentum will remain zero for all times, and the fact above then implies that the torques must sum to zero;
\begin{align}
  mgd - FD =0.
\end{align}
It follows that
\begin{align}
  F = \frac{d}{D}mg,
\end{align}
so the longer you make the part of the lever on the side opposite where the mass is hanging compared to the length of the part on the side where it is hanging, namely the smaller you make the ratio $d/D$, the lower the force $F$ you need to exert becomes.  In fact, no matter how large $m$ is, if the level is long enough, you can make the force you need to exert to hold it up as small as you wish!  This is the Law of the lever!
A: As you say, torque can be defined as $ r \times F $.  Then the law of the lever is derived simply from the condition that at static equilibrium all forces on the lever and all torques about the fulcrum sum to zero.
