I can't comment yet, so I'll provide a different way to explain the answer. But the point I most want to get across is that the internal reflections are never total; in fact, symmetry makes it so that TIR is impossible after sunlight enters a spherical raindrop. By definition, the sunlight enters at less than the critical angle (what gregsan described is called the rainbow angle, not the critical angle). TIR happens when it tries to exit at angles greater than the critical angle. Since symmetry requires it to strike the back of the drop at the same angle it entered, TIR can't happen.
Light doesn't strike a raindrop at just one angle of incidence, it strikes at all angles from 0 to 90 deg. The light that strikes at 0 deg (dead-center on the drop) does not refract at all: it goes straight in, reflects 180 deg, and comes straight back out in the direction of the sun. Light that strikes at an angle A=5 deg off-center, ends up being deflected about D=5 deg. As you increase the off-center angle A, the deflection D increases at first, but the rate at which it increases slows until the off-center angle is about 60 deg and the deflection is about 40 deg (the rainbow angle). Here the deflection angle function D(A) has a maximum. D decreases as A increases past this point.
As a result, the sunlight striking the drop reflects in a pattern that is much like the beam of an 80-deg wide flashlight, aimed toward the sun. You see this light reflected from raindrops everywhere between 0 deg (measured from the original sun's rays) to the rainbow angle. But here's the catch: the intensity of light at each deflection angle D is inversely proportional to D'(A). And since D(A) has a maximum at the rainbow angle, D'(A) is zero there.
So the light is, in theory at least, infinitely bright in an infintessimally-small band at the edge of the flashlight beam. You can't really see this edge, but it is still VERY bright close to it. Rainbows happen because the beam for each color has a different width, making this bright edge appear at a different place for each color.
And the green band is not pure green. It is a mix of all colors from green to red, it's just that the green is brighter. The mixture makes it a paler green than you'd see in a true spectrum. The same holds for every color after red. And there is no "inside edge" to the rainbow - it simply fades from violet to gray when no color dominates the mixture. This is why the "inside" is not dark - it isn't really inside at all, it is the gray band.
In Alexander's Band, outside the (primary) rainbow, no raindrops reflect light to you this way until you reach the secondary bow. Its flashlight beam is aimed away from the sun, and is about 260 deg wide. It "wraps" around the sky so you can see it while looking away from the sun. The colors are actually in the same order (violet=inside, red=outside), but you see the entire bow backwards (inside=up, outside=down) because of the wrapping.