TITLE: The $k$-skeleton of $n$-cube is homotopy equivalent to wedge of $k$-spheres.
QUESTION [6 upvotes]: The fact stated above can be obtained as follows: we just contract faces until all vertices come to one point. The question is how many spheres do we get?
One way to do this is to generalize the method given here: https://math.stackexchange.com/q/1174539 
Is there any other not so complicated way to do this?

REPLY [5 votes]: If a space is homotopy equivalent to the wedge of $m$ $k$-spheres, then
its Euler characteristic is $\chi=1+(-1)^km$, so we can find $m$ by computing $\chi$. In this case the cube structure gives an obvious CW-complex structure so that
$$\chi=\sum_{j=0}^k(-1)^j 2^{n-j}\binom n j.$$