Proving Wu's Formula for Steenrod Squares (Milnor & Stasheff, Problem 8-A) I am stuck at the following problem from chapter 8 of Characteristic Classes by Milnor & Stasheff.

Problem 8-A. It follows from 7.1 that the cohomology classes $\operatorname{Sq}^kw_m(\xi)$ can be expressed as a polynomial in $w_1(\xi), \dots, w_{m+k}(\xi)$. Prove Wu's explicit formula
$$\operatorname{Sq}^k(w_m) = w_kw_m + \binom{k-m}{1}w_{k-1}w_{m+1} + \cdots + \binom{k-m}{k}w_0w_{m+k},$$
where $\binom{x}{i} = x(x-1)\cdots(x-i+1)/i!$, as follows. If the formula is true for $\xi$, show that it is true for $\chi\times\gamma^1$. Thus by induction it is true for $\gamma^1\times\cdots\gamma^1$, and hence for all $\xi$.

Using the hint given, I have started to work out the following.
First, we prove the result for the line bundle $\gamma^1$ over $P^\infty$ as follows. Let $w_m \in H^m(P^\infty) \implies w_m=a^m$ where $a\in H^1(P^\infty)$. Claim:
$$\operatorname{Sq}^k(w_m) = \sum_{i = 0}^{k} {k-m \choose i} a^{k-i}a^{m+i}=\sum_{i = 0}^{k} {k-m \choose i}a^{k+m}$$ but I cannot proceed further. Using the total Steenrod square I was able to prove that $\operatorname{Sq}^k(w_m) =  {m \choose k} a^{k+m}$ but in vain.
Now suppose I have proved the above claim and it's true for the bundle $\xi$ then if we show that the result is true for $\xi \times\gamma^1$ then by induction argument we can say that it will be true for $({\gamma^1})^n$. Why is this enough to give the result for any n-plane bundle, I mean how exactly are we using the universality of $\gamma^n$.
Proof of second claim
\begin{align*}
&\ \operatorname{Sq}^k(w_m(\xi\times\gamma^1))\\ 
=&\ \operatorname{Sq}^k(w_m(\xi)\times 1+ w_{m-1}(\xi)\times w_1(\gamma^1))\\
=&\ \operatorname{Sq}^k(w_m(\xi)\times 1) + \operatorname{Sq}^k (w_{m-1}(\xi)\times w_1(\gamma^1))\\
=&\ \sum_{i +j=k}\operatorname{Sq}^i(w_m(\xi))\times \operatorname{Sq}^j(1)+ \operatorname{Sq}^i(w_{m-1}(\xi))\times \operatorname{Sq}^j(w_1(\gamma^1))\\
=&\ \operatorname{Sq}^k(w_m(\xi))\times 1 + \operatorname{Sq}^{k-1}(w_{m-1}(\xi))\times (w_1(\gamma^1))^2 + \operatorname{Sq}^{k}(w_{m-1}(\xi))\times w_1(\gamma^1)\\
=&\ \sum_{i = 0}^{k} {k-m \choose i} w_{k-i}(\xi)w_{m+i}(\xi)\times 1+\sum_{i = 0}^{k-1} {k-m \choose i} w_{k-i-1}(\xi)w_{m+i-1}(\xi)\times (w_1(\gamma^1))^2+\sum_{i = 0}^{k} {k-m+1 \choose i} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1).
\end{align*}
That's where I'm stuck as I'm not sure how to manipulate the third summand coefficients to get the desired thing as I need to have all the coefficients to be ${k-m \choose i}$. How can I get it?
Thanks & regards.
Edit 1: I think I figured out the use of Universality-
Since $(\gamma^1)^n$ is an n-plane bundle over ${(P^\infty)}^n$ so we have a map $f: {(P^\infty)}^n\to G_n(R^\infty) $covered by a bundle map $\gamma^1\to\gamma^n$ inducing an algebra monomorphism on the cohomology. Hence if the result is true for $(\gamma^1)^n$ it will be for $\gamma^n$ as we have that $$f^*(\operatorname{Sq}^k(w_m(\gamma^n)) = \operatorname{Sq}^k(w_m(f^*\gamma^n)) = \operatorname{Sq}^k(w_m(\gamma^1)).$$ Therefore we can again invoke the universality to say that the formula is true for any $n$-plane bundle. Any suggestions are welcome.
Edit 2: Here's how I manipulated the coefficients to complete the proof.
We can split the third summand of R.H.S using Pascal's identity to get
$$\sum_{i = 0}^{k} {k-m \choose i-1} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1)+\sum_{i = 0}^{k} {k-m\choose i} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1).$$
Now we can just replace $i-1$ by $i$ (as for $i=0$ in the former case it's just zero) in the first summand here to get
$$\sum_{i = 0}^{k} {k-m \choose i} w_{k-i-1}(\xi)w_{m+i}(\xi)\times w_1(\gamma^1)+\sum_{i = 0}^{k} {k-m\choose i} w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1).$$
Now substituting this above we get
\begin{align*}
&\ \operatorname{Sq}^k(w_m(\xi\times\gamma^1))\\
=&\ \sum_{i = 0}^{k} {k-m \choose i}[ w_{k-i}(\xi)w_{m+i}(\xi)\times 1 + w_{k-i-1}(\xi)w_{m+i-1}(\xi)\times (w_1(\gamma^1))^2\\ 
&\qquad\qquad\qquad + w_{k-i-1}(\xi)w_{m+i}(\xi)\times w_1(\gamma^1)+ w_{k-i}(\xi)w_{m+i-1}(\xi)\times w_1(\gamma^1)]\\
=&\ \sum_{i = 0}^{k} {k-m \choose i}(w_{k-i}(\xi)\times1+w_{k-i-1}(\xi)\times w_1(\gamma^1))\cup(w_{m+i}(\xi)\times1+w_{m+i-1}(\xi)\times w_1(\gamma^1))\\
=&\ \sum_{i = 0}^{k} {k-m \choose i}w_{k-i}(\xi\times\gamma^1)\cup w_{m+i}(\xi\times\gamma^1)
\end{align*}
concluding the proof.
Kindly help me with the first doubt.
 A: The two expressions for $\operatorname{Sq}^k(w_m(\gamma^1))$ are equal.
First note that
\begin{align*}
\binom{k - m}{i} &= \frac{(k-m)(k-m-1)\cdots(k - m - i + 1)}{i!}\\ 
&= (-1)^k\frac{(m-k)(m-k+1)\cdots(m - k + i -1)}{i!}\\
&= (-1)^k\binom{m - k + i - 1}{i}.
\end{align*}
In particular, $\displaystyle\binom{k-m}{i} = \binom{m-k+i-1}{i}$ modulo $2$.
Now note that by repeated application of Pascal's rule we have
\begin{align*}
\binom{m}{k} =& \binom{m-1}{k} + \binom{m-1}{k-1}\\ 
=& \binom{m-1}{k} + \binom{m-2}{k-1} + \binom{m-2}{k-2}\\ 
 \vdots & \\
=& \binom{m-1}{k} + \binom{m-2}{k-1} + \dots + \binom{m-k}{1} + \binom{m-k}{0}\\
=& \binom{m-1}{k} + \binom{m-2}{k-1} + \dots + \binom{m-k}{1} + \binom{m-k-1}{0}\\
=& \sum_{i = 0}^k\binom{m-i-1}{k-i}\\
=& \sum_{i = 0}^k\binom{m-k+i-1}{i}.
\end{align*}
Therefore
$$\sum_{i=0}^k\binom{k-m}{i}a^{k+m} = \sum_{i=0}^k\binom{m-k+i-1}{i}a^{k+m} = \binom{m}{k}a^{k+m}.$$

Due to the modulo $2$ equality of binomial coefficients mentioned above, Wu's formula is more commonly written as
$$\operatorname{Sq}^k(w_m) = \sum_{i=0}^k\binom{m-k+i-1}{i}w_{k-i}w_{m+i}.$$
