Why is the circumferential component of the magnetic field around a solenoid zero? According to Introduction to Electrodynamics by Griffiths, the reason is "$B_{\phi}$ would be constant around an Amperian loop concentric with the solenoid, and hence $\int{\vec{B}  \cdot d\vec{l}}=B_{\phi}(2\pi s) = \mu_0 I_{enc}=0$ since the loop encloses no current."
My questions: 
- Why would $B_{\phi}$ be constant along the circumferential direction?
- Why would the loop enclose no current? (surely, it encloses a small part of at least one wire, so I imagine for that reason that it encloses current).
And, additionally, an explanation of why the field inside is in the $\hat{z}$ direction inside the solenoid. 
An explanation of why a rectangular Amperian loop is useful would also be helpful.
 A: 
Why would Bϕ be constant along the circumferential direction?

Because of cylindral symmetry; each point in the (circular) Amperian loop is "positionally equivalent" relative to the solenoid (the distance to the solenoid wires is the same and the current in the solenoid does not change along the circumferencial direction).

Why would the loop enclose no current? (surely, it encloses a small part of at least one wire, so I imagine for that reason that it encloses current).

It is an approximation.

And, additionally, an explanation of why the field inside is in the z^ direction inside the solenoid.

Right hand rule, implicit in Ampère's law. Sum the contributions of each wire to $\vec{B}$ at a point inside the solenoid.

An explanation of why a rectangular Amperian loop is useful would also be helpful.

The Amperian loop shape is usually choosen taking into account the symmetry of the problem, so that it becomes easy to calculate $B$ by using the integral form of Ampère's law - usually, $B$ is constant along the path, except maybe in some parts where it may be null, as it is the case. A similar approach is used with Gauss' law, but instead of a path, a useful surface - along which $E=cte$ - is chosen.
