TITLE: The "Wu formula" and Steenrod algebras
QUESTION [17 upvotes]: The Wikipedia page on Stiefel-Whitney classes includes the following paragraph:

Over the Steenrod algebra, the Stiefel–Whitney classes of a smooth
  manifold (defined as the Stiefel–Whitney classes of its tangent
  bundle) are generated by those of the form $w_{2^i}$ . In particular, the
  Stiefel–Whitney classes satisfy the Wu formula, named for Wu
  Wenjun:
  $\text{Sq}^i(w_j)=\sum_{t=0}^i {j+t-i-1 \choose t} w_{i-t}w_{j+t}.$

This seems to suggest the first sentence follows from the second, but the reference provided (to May's A Concise Course in Algebraic Topology) doesn't mention this. Personally, I can't see the connection. Is there some well-known lemma I'm missing?

REPLY [12 votes]: By expanding out the Wu formula and moving some terms over to the other side, you find that
$$
{j - 1 \choose i} w_{i+j} = Sq^i(w_j) + \sum_{t = 0}^{i-1} {j+t-i-1 \choose t} w_{i-t}w_{j+t}.
$$
If the binomial coefficient on the left is nonzero, this allows you to write $w_{i+j}$ as a sum of terms generated by lower Stiefel-Whitney classes.  So for a fixed $n$, this becomes a question as to whether there exist $i$, $j$ such that $i+j = n$ and $j-1 \choose i$ is nonzero.
If $n$ is not a power of $2$, we can write it as $2^k + i$ where $2^k > i$.  Then $2^k - 1 \choose i$ is nonzero.  (By Lucas' theorem, the binomial coefficient nonzero if and only if, in binary, the nonzero digits of $i$ are also nonzero digits of $j-1$.  The number $2^k-1$ has all of its binary digits nonzero, and $i$ is less than or equal to it.)
If $i+j$ is a power of two, then $i + (j - 1)$ has no zeros in its binary expansion, which means that $i$ and $j-1$ can't share any nonzero binary digits; there can be no carries when adding $i$ and $(j-1)$ together.