VC dimension of monotone disjunctions of length k over n variables?
Yes.  This bound follows by a little bit of straightforward manipulation:

$${n \choose k} = {n \times (n-1) \times \dots \times (n-k+1) \over k!} \le {n \times n \times \dots \times n \over 1} = n^k.$$

Now taking the logarithm of both sides, we see that

$$ \lg {n \choose k} \le \lg(n^k) = k \lg n.$$

The bound can also be derived from bounds listed in Wikipedia: it follows from the fact that ${n \choose k} \le n^k/k!$.  Check out Wikipedia in the future -- it's often helpful!