This is part of problem 11-E in Milnor and Stasheff's Characterisitic Classes, which reads:
Prove the following version of Wu's formula. Let $$ \overline{Sq}:H^\Pi(M)\rightarrow H^\Pi(M) $$ be the inverse of the ring automorphism $Sq$. Show that the dual Stiefel-Whitney classes $\overline{w}_i(TM)$ are determined by the formula $$ \langle \overline{Sq}(x),\mu\rangle = \langle \overline{w}\smile x,\mu\rangle $$ which holds for every cohomology class $x$. Show that $\overline{w}_n=0$. If $n$ is not a power of $2$, show that $\overline{w}_{n-1}=0$.
I am able to prove the formula, and I see why $\overline{w}_n$ vanishes, but I do not see what not being a power of 2 has to do with the vanishing of $\overline{w}_{n-1}$. Any hints would be much appreciated.
It is enough to prove that for each $x\in H^1(M,\mathbb{Z}_2)$ we have $(\overline{Sq}(x),[M])=0$ and the conclusion holds by Poincaré duality. We claim that $$\overline{Sq}(x)=x+x^2+x^4+\dots,$$ which immediately implies $(\overline{Sq}(x),[M])=0$ as long as $n$ is not a power of $2$. In fact, notice that $$Sq(x^{2^l})=Sq(x)^{2^l}=(x+x^2)^{2^l},$$ and therefore $Sq^i(x^{2^l})=0$ for $i\not=0,2^l$. We have \begin{align*} Sq(x+x^2+x^4+\dots)&=Sq^0(\sum_{l=0}^{\infty}x^{2^{l}})+\sum_{l=0}^{\infty}Sq^{2^l}(\sum_{i=0}^{\infty}x^{2^{i}})\\ &=\sum_{l=0}^{\infty}x^{2^{l}}+\sum_{l=0}^{\infty}Sq^{2^l}(x^{2^l})\\ &=\sum_{l=0}^{\infty}x^{2^{l}}+\sum_{l=0}^{\infty}x^{2^{l+1}}\\ &=x. \end{align*}Here all the expressions are finite sum and we don't have to worry about indices.