Given a loop in Sn, let p be a point not on the loop, and let q be the point on the other side of the sphere, opposite to p. Linearly drive the entire sphere away from p and towards q along great circles. This is a homotopy that shrinks the loop to a point.
To be technically correct, q should be set to the base point, which could be right next to p. One can still build a homotopy that pushes the path out away from p, all the way around the sphere, and onto q. Picture a small circle about p, expanding and stretching away from p, until it becomes a great circle equidistant between p and q. By symmetry, the circle swings across the other side of the sphere and closes in on q, pushing and compressing the embedded loop ahead of it. In a 3 sphere, picture a 2 sphere expanding outward from p, becoming a great sphere between p and q, then swinging around and closing in on q. I'll let you derive the formula for this homotopy in n dimensions yourself.
As a generalization, the above homotopy pushes, and crunches, the image of any domain (not just a loop) onto the point q, provided the image misses the point p.
Of course the above homotopy won't work unless the loop can be bounded away from p. In other words, p has to be part of a neighborhood that is free to expand, pushing the loop towards q. The sphere is a regular space, and the loop and p are both closed, hence they can be separated in disjoint open sets. This establishes the neighborhood about p, and we're on our way.
But there's still a problem. What if the loop covers the entire sphere, so that there is no such point p? After all, space filling curves do exist. In what follows, we will "homotopy" the troublesome curve, so that it avoids the point p.
Draw a small circle on the surface of the sphere, such that the base point q lies outside the circle. Place p at the center of the circle. The disk, defined by this circle, is a closed set, which means the portion of the loop that runs through the circle has a closed preimage C in I.
Let x be a point of C that maps to the interior of the circle. Let a be the least upper bound of points in I, below x, that do not map inside the circle. We know that 0 < a < x, and a maps to the border of the circle. Similarly, let b be the greatest lower bound of points above x that do not map inside the circle. Now the closed interval [a,b] maps into the circle, with a and b mapping to the border, and the rest inside.
Assume u and v are the images of a and b respectively. Let y be any point on the circle other than u and v. The path from u to v is compact in a complete metric space, hence it is closed. If it approaches y it contains y, which is a contradiction. Hence there is a neighborhood about y that is free to expand, pushing the path away. construct a homotopy that pushes the path from u to v away from y and towards the opposite side of the circle. this looks like a tiny bubble that starts at y and expands linearly with time, filling the interior of the circle and pushing the path ahead of it. You can push the path all the way to the far side of the circle if you like, but for our purposes, we only need push the path past p, so that p is not part of the path.
Note that u and v remain fixed throughout the homotopy; we are only warping the interior of the circle. The rest of the path, outside the circle, is unaffected by this homotopy.
build a homotopy similar to the above whenever a point x maps into the circle. This may happen infinitely often, so we need to run all these homotopies in parallel. Every section of the path that wiggles about inside the circle is pushed to one side or another, so that it misses p. The new path, homotopic to the original, misses p, and that completes the proof.
All this works in higher dimensions. Given an n sphere, draw an n-1 sphere on the surface, missing the base point q, and place p at its center. A partial path from u to v can be pushed to one side by a homotopy expanding inward from y. Do this for all the subpaths running through the n-1 sphere, and the loop misses p, whence it can be pushed onto q.