An $n$-hypercube is both $n$-vertex- and -edge-connected I have come across some proofs for this, most of which take the $n$-disjoint paths approach. I came up with the following induction idea and I'd like to ask whether or not you find it sound.
The hypothesis holds for the 2-hypercube.
Suppose that the hypothesis holds for the ($n-1$)-hypercube. Then, we note that the $n$-hypercube is formed by two copies Q and Q' of the ($n-1$)-hypercube with a perfect matching between their vertices. Then, for Q (or Q') to be disconnected we have to remove $n-1$ vertices from it. So one vertex in Q is isolated and to disconnect the  $n$-hypercube we need to also remove its pair from Q'.
 A: Based on the above comments, I am posting my full proof, hoping it's correct.
The base is that $Q_2$ is 2-connected.
Then we observe that, $Q_n$ is an $n$-regular graph (easily proved by induction), so we can disconnect it if we remove the $n$ neighbors of a random vertex. So the connectivity $k$ is equal or less than $n$.
The induction hypothesis is that the hypercube $Q_{n-1}$ is $n-1$-connected, thus so are the hypercubes $Q$ and $Q'$ as defined above.
Let $S$ be a minimum separator of $Q_n$. We have two cases. 
Case 1. $S$ leaves both $Q$ and $Q'$ connected, so in order for $Q_n$ to disconnect, we have to remove ALL the edges connecting $Q$ and $Q'$, that is a vertex at one end of each edge. We know from the definition of hypercubes that both $Q$ and $Q'$ have $2^{n-1}$ vertices and thus, $S$ will have at least $2^{n-1}\ge n$ vertices. 
Case 2. If $S$ disconnects one of the small hypercubes, say $Q$ wlog, then it has at least $n-1$ vertices (because it is $n-1$-connected by the induction hypothesis) and we also have to remove at least one more vertex from $Q'$ to disconnect $Q_n$. So again, $S$ has at least $n$ vertices, which completes the proof because of our earlier observation.
NOTE: One can now easily prove that $Q_n$ is also $n$-edge-connected, since it is $n$-regular and the following holds every graph $G$:
$k(G) \le \lambda (G) \le \delta (G)$
where $k$ is the vertex-connectivity, $\lambda$ is the edge-connectivity and $\delta$ is the minimum degree of G.
