What is the intuition for higher homotopy groups not vanishing?  
The homotopy groups of the spheres $S^n$ (see Wikipedia) vanish for the circle $S^1$ as, naively speaking, there are not higher order holes to be grasped by higher order homotopy groups. This intuitions already breaks down for the two sphere $S^2$, e.g. $\pi_3(S^2)$ is non-trivial because of the Hopf fibration. This non-triviality seems to keep on going for all the higher spheres $S^n$. What makes $S^1$ so fundamentally different? 
 A: One explanation follows from the fact that if $X$ is a space and $\tilde X$ is its universal cover, then for $i\geq 2$ we have $\pi_i X \cong \pi_i \tilde X$. 
Then you can just observe that the universal cover of $S^1$ is $\mathbb R$ (which is contractible and hence has vanishing higher homotopy groups), while for $n > 1$, the universal cover of $S^n$ is just $S^n$ itself (it is simply-connected, so you can just take the identity as a covering map).
A: Iʼm no homotopy theorist, but I have a little scrap-let of intuition that may be helpful. I havenʼt studied the underlying concepts enough to go into much meaningful detail here. Maybe someone else can expand on this; maybe my thoughts are utterly wrong. But without further ado…
I'm going to compare the 1-sphere and the 2-sphere by describing them in alternation, one paragraph each. 
If you want to trace out a 1-sphere, one way to do that is to take a 0-sphere (a pair of points), anchor one of the points, and move the other point in a loop, starting and ending at that anchor point. By doing that, you've traced out a path within the 1-sphere, and that path is the generator of the space. 
Likewise, if you want to trace out a 2-sphere, you can take a 1-sphere, anchor one of the points, and move the opposite point in a loop, starting and ending at that anchor point. By doing that, you've traced out a homotopy (a 2-path) within the 2-sphere, and that homotopy is the generator of the space.
Of course, the 1-sphere has more loops (paths from the anchor point to itself) than just the generator. There's a trivial loop, and you can also reverse loops and compose them. These are, of course, simply the operations of a group, and the homotopy group $\pi_1(S^1)$ describes how these operations work.
Likewise, the 2-sphere has more 2-loops (homotopies from the trivial loop on the anchor point, to itself) than just the generator. You have the group operations again, described by the homotopy group $\pi_2(S^2)$.
With the 1-sphere, the group operations "tell the whole story". Up to homotopy, there are no more loops than those created by the group operations.
With the 2-sphere, the group operations no longer tell the whole story. The generator we identified consists of taking a point and moving it in a circle in a particular direction. The group operations allow us to move in the opposite direction. But it's also possible to move in a perpendicular direction, or sideways, or in any of $S^1$-many directions. So in order to tell the whole story, we need additional homotopy groups.

Of course, the question I'm failing to answer is: how, exactly, do additional homotopy groups tell the rest of the story? I don't really know. But hopefully I've given some motivation how, unlike the loop space of $S^1$, the loop space of $S^2$ is itself an interesting space; and the loop space of that space is an interesting space, and so on. 
A: So far the discusion mostly focused on a geometric explanation, I'd like to mention the algebraic one as well:
One way to formulate it involves the delooping machinery: up to delooping, $\mathbb{S}^n$ corresponds to the free group like $E_n$-algebra on one generator.

Small recall: the usual delooping machinery say that the looping/delooping construction induce an equivalence between pointed spaces $X$ such that $\pi_k X =0$ for all $k<n$ and group like $E_n$-algberas. Through that correspondence $\mathbb{S}^n$ corresponds to the free group like $E_n$-algebra on one generators, as the looping/delooping adjunction justs shift the $\pi_n$ you can describe the homotopy group of sphere as shifted homotopy group of these free group like $E_n$-algebra. A more explicit way to say all this is that the free group like $E_n$-algebra is $\Omega^n \mathbb{S}^n$.

Now, when you construct the free $E_1$-algebra, there is not much you can do: you only have one way to multiply elements and the free $E_1$-algebra on one generator is just the monoid $\mathbb{N}$ (and $\mathbb{Z}$ for the group like one).
But for $E_n$ you get $n$ "compatible" (in a homotopy theoretic sense) way to multiply the elements and all the higher elements in the homotopy group comes from the interaction (the coherence law, that are given by homotopies) between these multiplications, for example the free $E_2$-algebra has all the braid groups appearing as its various $\pi_1$ due to that, and the free group like $E_2$-algebra become too complicated to described (well... it is $\Omega^2 \mathbb{S}^2$ ).
So the difference is that for $n=1$ no such interaction is happening because to get interaction you need at least two compatible multiplication.
Alternatively to the delooping machinery, one can (somehow equivalently) think of spaces as $\infty$-groupoids and as the $n$-sphere as the $\infty$-groupoid freely generated by a cell in dimension $n$. The discussion is pretty much the same except that now the $n$ "compatible" multiplication are simply the compositions in direction $k$ for $k$ from $0$ to $n-1$.
Edit: Here is how you get a non trivial element of $\pi_3(\mathbb{S}^2)$, in the second perspective. I'm using an unspecified model of weak $\infty$-groupoid, and applying freely the operation of strict $\infty$-categories to give a feel of how it works, this is not mean't to be formal (but it is formalizable in any algebraic model of weak $\infty$-groupoid, or in Hott )
Given a two cells $u$ and $v$ whose source and target is a (weak) identity, the usual Eckman Hilton argument (so the typical example of interaction between $\#_0$ and $\#_1$ as I mentioned above) gives an isomorphism $\theta_{u,v} : u \#_0 v \simeq v \#_0 u$.
If $e$ is the generating 2-cell of the 2sphere then this gives an isomorphisms $\theta_{e,e}: e \#_0 e \simeq e \#_0 e $ 
taking $e^*$ a $0$ inverse of $e$, one has that $e^* \#_0 \theta_{e,e} \# e^*$ is a 3-cell whose source and target are (up to the coherence isomorphism expressing that $e$ and $e^*$ are inverse) identities, so it gives an element of $\pi_3(\mathbb{S}^2)$, which is non-zero by a universality argument. I'm convince it is a generator (so either the Hopf fibration or its opposite depending which way you have rotated the 'Eckman-Hilton clock') but I don't know how to prove it using only this type of tools. 
A: Homotopy groups of spheres correspond to framed submanifolds of Euclidean space through the Pontrjagin-Thom construction.  For example, the Hopf map corresponds to a circle in $\mathbb{R}^3$ framed “with a twist”.  The homotopy groups of $S^1$ thus correspond to framed codimension one submanifolds.  But such are canonically framed and all bound, so there are no interesting/ non-trivial examples.
A: In a sense, failure of the the higher homotopy groups of $S^n$ to be trivial, $n>1$, is due to them not representing singular cohomology. If the higher homotopy groups were trivial, all spheres would be Eilenberg-MacLane spaces and would represent cohomology. For most spheres, this failure to represent cohomology can be seen because they are not loop spaces which in turn is because they are not groups. 
A: Here's a very simple argument as to why not all of the spheres have higher homotopy groups vanish. Higher homotopy groups vanishing (along with the lower ones) is equivalent to the sphere representing n-dimensional cohomology. By Yoneda, natural transformations between functors correspond to maps between representing objects. Hence, natural transformations between nth cohomology with integral coefficients and mth cohomology with integral coefficients correspond to maps $S^n \rightarrow S^m$ if these both represent singular cohomology. However, there are nontrivial cohomology operations (say for example, reduction mod p followed by the integral Bockstein), and so this implies that we have a map between the spheres representing them. However, this would imply that the higher homotopy groups of this sphere do not vanish, and so we get a contradiction.
