Number of automorphisms of n-dimensional hypercube graph. Hypercube graph definition from Wikipedia:

In graph theory, the hypercube graph $Q_n $is the graph formed from the vertices and edges of an n-dimensional hypercube. 

My teacher gave the number $2^n n!$ but I don't know how to compute it.
First I tried prove it by induction on $n$. Base case for $n=2$ is simple. However, I ran into some trouble for higher dimensions. It's really hard even to visualize 4-dimensional space.
Then I tried to prove it by the Stabilizer-Orbit theorem. Also failed for high dimensions.  
This seems like a really elementary question. Need a little help.
 A: Hint 1
Given a vertex $v\in V(Q_n)$. An automorphism of $Q_n$ is uniquely determined by the images of $v$ and all its neighbors $w\in N(v)$. Can you see why and how this gives $2^nn!$ ?

Hint 2
I assume the $n$-dimensional hypercube graph $Q_n=(V,E)$ is given by the vertex set $V=\{0,1\}^n$ of all binary sequences of length $n$, and $\{v,w\}\in E$ if and only if the two sequences $v$ and $w$ differ in exactly one element.

Theorem. The automorphisms of $Q_n$ are exactly the maps $$v\mapsto (v\circ\sigma)\oplus s$$ for vertices $s\in V$ and permutations $\sigma:\{1,...,n\}\to\{1,...,n\}$. Here $\oplus$ denotes the element-wise XOR operation. Note that every vertex can be considered as a map $\{1,...,n\}\to\{0,1\}$, hence the concatenation $\circ$ is meaningful.

Try to prove this. First show that this map is indeed an automorphism. Then (and this might be the harder part) show that any automorphism is of this form, e.g. by showing that every automorphism has an inverse of this form.
Once you have achieved this, observe that there are $2^n$ vertices $s\in V$ and $n!$ permutations $\sigma$. This gives a total of $2^nn!$ automorphisms.

Still another way. Prove

Theorem. Let $v\in V$ be a vertex in $Q_n$ and $N(v)$ its neighborhood. A vertex $w\in V$ is uniquely determined by its distances $d(w,u),u\in N(v)$.

You can use this to make Hint 1 into a rigorous proof.
A: Here is an intuitive (non-rigorous) reasoning why the answer might be $2^nn!$.
Lets say your $n$-dimensional hypercube is given as the set $[0,1]^n$. An automorphism is a transformation of the cube after which it looks exactly like before. In the case of the hypercube, these transformations are all rigid rotations and reflections (this would need a proof when done rigirously).
So think about this hypercube as a small piece of [material of your choice] sitting in the corner of the positive orthant $\Bbb R^{n+}$. Now take it into your hand, rotate it a bit, and place it back into the corner. What could have changed? For example: the corner which touches the orthant's corner $(0,...,0)$ has changed. There are $2^n$ corners of the hypercube, so there are already $2^n$ choices which one to place there.
What other options do we have? Lets say it is the corner $v$ of the hypercube which gets placed in the orthant's corner after the transformation. There are $n$ edges of the hypercube which start in $v$. And now we have the option to place any of these edges on any of the orthant's axes. E.g. for a $3$D-cube, 
\begin{align}
\text{place the hypercube's $x$-edge on the orthant's $y$-axis,}\\
\text{place the hypercube's $y$-edge on the orthant's $z$-axis,}\\
\text{place the hypercube's $z$-edge on the orthant's $x$-axis.}
\end{align}
Geometrically, this is a rotation of the $3$-cube. With rotations and reflections you can achieve any possible permutations of the $\{x,y,z\}$-edges onto the $\{x,y,z\}$-axes. This is still true in higher dimensions (but not obvious). Any permutation is possible and can be achieved by a series of rotations and reflections. So for any transformation which moves the corner $v$ into the orthant's origin, there are still $n!$ (that is the number of permutations) choices on how to place the edges on the axes. Together with the $2^n$ choices of the corner $v$ we have $2^nn!$ choices for the automorphisms.
