= Solution
Let $i:F\hookrightarrow E$. A pair
$$
(x,y)\in H^r(F;\mathbb F_2)\times H^{r+1}(B;\mathbb F_2)
$$
is <transgressive pair> when, in the long exact sequence of the pair $(E,F)$,
$$
\delta x=\pi^*y\in H^{r+1}(E,F;\mathbb F_2),
$$
where $H^{r+1}(B,b_0)$ is identified with reduced cohomology. In the <Serre spectral sequence>, this says that $x$ survives to the transgression and
$$
d_{r+1}(x)=y
$$
under the edge identifications, modulo the usual earlier-differential indeterminacy.
The <Kudo transgression theorem> says that if $(x,y)$ is transgressive and $0\leq j\leq r$, then
$$
(\operatorname{Sq}^j x,\operatorname{Sq}^j y)
$$
is transgressive. To prove it, use relative <Steenrod squares>. Naturality gives
$$
\operatorname{Sq}^j(\pi^*y)=\pi^*(\operatorname{Sq}^j y),
$$
and stability, equivalently compatibility with the suspension isomorphism, makes squares commute with the connecting map:
$$
\operatorname{Sq}^j(\delta x)=\delta(\operatorname{Sq}^j x).
$$
Applying $\operatorname{Sq}^j$ to $\delta x=\pi^*y$ proves the theorem. The properties used are naturality, stability, additivity, and the instability conditions $\operatorname{Sq}^jz=0$ for $j>|z|$ and $\operatorname{Sq}^{|z|}z=z^2$.
Let $x\in H^1(\mathbb{RP}^\infty;\mathbb F_2)$ be the generator. Instability gives
$$
\operatorname{Sq}(x)=\operatorname{Sq}^0x+\operatorname{Sq}^1x=x+x^2.
$$
The <Cartan formula> says the total square is multiplicative, so
$$
\operatorname{Sq}(x^k)=(x+x^2)^k
=\sum_{j=0}^k\binom{k}{j}x^{k+j}.
$$
Comparing components yields the complete formula
$$
\boxed{\operatorname{Sq}^j(x^k)=\binom{k}{j}x^{k+j}},
$$
with the binomial coefficient reduced modulo two.
Finally, $\operatorname{Sq}^1$ is the <Bockstein homomorphism> associated with
$$
0\longrightarrow\mathbb Z/2\longrightarrow\mathbb Z/4
\longrightarrow\mathbb Z/2\longrightarrow0.
$$
If a mod-two cocycle representing $y$ is lifted to an integral cochain $A$, write $dA=2B$. Then $B$ modulo two represents $\operatorname{Sq}^1y$. But $dB=0$, because integral cochains are torsion-free and $2dB=d^2A=0$. Thus $B$ itself is a cocycle lift, so its Bockstein vanishes. Therefore
$$
\operatorname{Sq}^1\operatorname{Sq}^1(y)=0
$$
for every space $Y$ and every $y\in H^*(Y;\mathbb F_2)$.
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