I would appreciate knowing if I am correct below.
I assume that the cohomology group $H^i(\Bbb P^n,\Bbb Z/2\Bbb Z)$ is cyclic order $2$ for $0\leq i \leq n$ and is zero for higher values of $i$. Letting $a$ be the generator of $H^1(\Bbb P^n,\Bbb Z/2\Bbb Z)$ and $i$-fold cup product $a^i$ generated the $i$-th cohomology.
I then want to show that $w(\gamma_n^1)=1+a$
So we can obtain elements of $\Bbb P^1$ by taking $\Bbb S^1\to \Bbb P^1$ via $$(\cos(\theta),\sin(\theta))\mapsto \{\pm(\cos(\theta),\sin(\theta))\}$$ And then we can embed this into $\Bbb P^n$ via $$\{\pm(\cos(\theta),\sin(\theta))\}\mapsto \{\pm(\cos(\theta),\sin(\theta),0,0,\cdots,0)\}$$ where here there are $n+1$ terms, right. (i.e. $f:B(\gamma_1^1)\to B(\gamma_n^1)$ is given as above)
This is then covered by a bundle map: $$E(\gamma_1^1)\to E(\gamma_n^1)$$ given by the line $L\in \Bbb R^2$ through $\pm x$ in the domain, and $L_2\in \Bbb R^{n+1}$ the line through $\pm x$ in the image in $$\{\pm x\} \times L\mapsto\{\pm x\} \times L_2 \in B(\gamma_n^1)\times \Bbb R^{n+1} $$
This thus allows us to use naturality right? I.e. now we can say $$f^* w_1(\gamma_n^1)=w_1(\gamma_1^1)$$
And regardless of what $f^*$ is, I know it is a homomorphism, and so since $w_1(\gamma_1^1)$ is not trivial, it follows that $w_1(\gamma_n^1)$ is not trivial(else it would be mapped to $0$). Then $\gamma_n^1$ is a $1$-plane bundle right? So we are done by rank axiom right?