Let's look at the simplest cases: you shine your beam at the point on the wall closest to you, or at the point on the wall farthest from you, or you're at the center of the sphere. In any of these cases, the beam will bounce right back and hit you. There.
Now let's assume that you don't do any of that.
A good first step is to simplify the problem.
The first simplification I will make is this: there exists a plane containing the entirety of the path of your light beam; that plane also contains you and the center of the sphere.
The light will hit the side of your sphere and reflect off of it based on the angle of incidence of the beam on the plane tangent to the sphere at that point. Since that tangent plane is normal to a line drawn from the center of the sphere to the point of incidence, you end up with the first reflection of the beam remaining in the plane with you, the point of incidence and the center of the sphere. Now if you pick a point on your beam's path between its first and second points of reflection and think of it as a new source, you know that the second reflection of the beam will end up on a plane which includes your first beam reflection and the center of the sphere, for the same reason that your first beam reflection is in the same plane as your original beam and the center of the sphere. Since these two planes share your first reflected beam and the center of the sphere, you know that they must be the same plane. By induction, you'll find that every segment of your beam exists on one plane, which also includes you and the center of the sphere.
So the simpler problem, which will have the same answer, is this: If you shine a beam of light at the wall of a mirrored circular enclosure from any point inside the enclosure, will the light hit you?
Well, let's again look at the simplest case remaining: your first angle of incidence is a rational multiple of 2π.
In this case, your beam will hit you. This is because its path loops, which it will not do otherwise.
In this and the remaining cases, you will find that every time your beam reflects, the angle of incidence is the same. You can check this by drawing any chord on a circle and drawing line segments from the endpoints to the circle's center. Since the resulting triangle is an isosceles, its two base angles are the same. When you shine your beam in the circular enclosure, every beam segment (after the first) is a chord between two points on your enclosure's wall, and the two base angles on its isosceles are the angle of incidence on one reflection and the angle of reflection on another; since angle of incidence equals angle of reflection, every angle of incidence must be equal.
Think of an arrow that points in the direction of motion of your light beam. The arrow would rotate by twice your angle of incidence each time your beam reflected; it would either always go clockwise or always go counterclockwise. After a full cycle is completed on a looped path, the arrow will be pointing in the same direction it was when it started. Thus, for the beam's path to loop, its total rotation—which would be an integer times twice your angle of incidence—would have to be an integer multiple of 2π; thus, the angle of incidence would have to be a rational multiple of 2π. So if your angle of incidence is not a rational multiple of 2π, the path will not loop.
But if your angle of incidence is a rational multiple of 2π, the path loops. A little more geometry shows that every beam segment (again barring the first) is the same length, no matter what that initial angle is. The length of a chord on a circle can be determined entirely by the radius of the circle and the angle between the chord and a line segment drawn from the center of the circle to an endpoint on the chord, so the constant angle of incidence directly implies that every beam segment has the same length.
Given a circle and the fact that a chord of a given length is drawn in a given direction, there are only two possible chords for those criteria, one of which has the center of the circle to its left (by the drawing direction) and the other of which has the center of the circle on its right. The center of the circle will always be on the same side of your beam (from the beam's perspective). Since your beam will have multiple segments that go in the same direction due specifically to your angle of incidence, it follows that your beam will traverse the same chords repeatedly in the same direction—which means that your path will be looped. So in this case, yes, your beam will hit you.
So in the remaining cases, your initial angle of incidence is not a rational multiple of 2π, and so your path will not loop.
Let's start with a case in which your beam will certainly not hit you: your starting beam is perpendicular to the line containing you and the center of the circle.
In this case, the distance between you and the center of the circle is as close as any of your beams will pass to the center of the circle. Each beam segment will pass this close to the center of the circle exactly once, and that point will be its midpoint; thus, since a chord can be defined entirely by the location of its midpoint in the circle, only the chord on which your initial beam segment lies will have you on it. Since your path does not loop, your beam will not traverse this chord again; thus, your beam will not hit you.
The final cases are those in which your angle of incidence is not a rational multiple of 2π and your beam is not perpendicular to the line containing you and the center of your circle. Your beam's path will not loop.
Given a point in a circle and a length, if that length is greater than the length of the shortest chord that goes through that point and shorter than the diameter of the circle. then there are exactly two chords of that length that go through that point. So given the length that your beam segments take and your location in the circle, if your beam hits you, it hits you on the chord of that length that goes through you, but does not contain your initial beam segment, because your path will not be retraced. If there exist nonnegative m, n such that the angle between these two chords (the one that the line from the circle's center to you goes through) is equal to (m*2*the angle of incidence) minus 2πn, then your beam will include that chord and hit you; otherwise, it never will. This is because after m reflections, your beam's direction will have rotated by (m*2*the angle of incidence), which will be 2πn plus the difference in angle between your initial and final beam segments. This will select the final beam segment as the chord that will hit you. Otherwise, your beam will never traverse that chord.
I might have been flaky on a few details, but I hope this answers your question.