Can $S^n$ be defined without reference to an $n+1$ dimensional embedding space? It seems like the most common definition is the set of all vectors in $\mathbb{R}^{n+1}$ which are unit length. Is this a necessary definition?
 A: If you just want something homeomorphic to $S^n$, you can use the one-point compactification of an open ball in $\Bbb R^n$.
A: Up to homeomorphism, $S^n$ is a triangulated space whose gluing data is that of the $n+1$-dimensional simplex, but with the highest-dimensional simplex removed. (So, for example, you can deform a tetrahedron to a sphere after you remove the stipulation that it has to stay in the shape of a tetrahedron.)
A: For $n\ge 2$ it is the unique (up to homeomorphism) compact, simply connected Riemannian manifold which is strictly $1/4$-pinched (Berger, Klingenberg). 
Brendle and Schoen even proved a stronger result: a compact, simply connected manifold $M$ whose sectional
curvatures all lie in the interval $(1, 4]$ is necessarily diffeomorphic to
the sphere $S^n$, see their paper Manifolds with 1/4-pinched Curvature are Space Forms.
A: If you are interested in topology, then you can naturally construct the spheres inductively with no reference to any embedding. You have to believe that suspension is a natural operation, though. But it is.
Then $S^0$ is a discrete space with two points, and $S^{n+1}$ is the suspension of $S^n$.
If you are only interested in homotopy, then the spheres appear spontaneously a whole lot ; for instance, $S^n$ is the (pointed) space that represents the functor $\pi_n$ (though it may take a little leap of faith to really believe that the homotopy groups are interesting without knowing beforehand their relation with spheres, but you can indeed make a very strong case for that).
