algebraic topology – Show element of fundamental group is nontrivial

I’m learning cohomology, and I’d like to show the following:

Let $i:mathbb RP^1tomathbb RP^n$ be the usual embedding taking $(x_0,x_1)mapsto(x_0,x_1,0,dots,0)$, where $nge2$. Further, let $v:S^1tomathbb RP^1$ be the fibration $(x_0,x_1)mapsto(x_0,x_1)$. Show that $(icirc v)$ is a nontrivial element of $pi_1(mathbb RP^n,*)$.

I tried proof by contradiction: Suppose $icirc v$ is homotopic to the constant map $c$ which sends everything to $*$. Then these induce the same maps in cohomology: $v^*circ i^*=(icirc v)^*=c^*$. Recall that $H^q(mathbb RP^m;mathbb Z/2mathbb Z)=mathbb Z/2mathbb Z$ for all $0le qle m$. Let $Omega_nin H^1(mathbb RP^n;mathbb Z/2mathbb Z)$ be the nonzero element, and similarly define $Omega_1in H^1(mathbb RP^1;mathbb Z/2mathbb Z)$. Then I already know that $i^*(Omega_n)=Omega_1$. The Gysin sequence shows that $v^*(Omega_1)=0$.

I wanted to show that $c^*(Omega_1)$ is nonzero. But I feel like I don’t quite understand cohomology groups/rings enough. If we write $Omega_n=text{cls}~omega_n$, where $omega_nin Z^1(mathbb RP^n;2)$ is a homomorphism $C_1(mathbb RP^n)tomathbb Z/2mathbb Z$, then the goal is to show that $$omega_nc_#:C_1(S^1)to C_1(mathbb RP^n)tomathbb Z/2mathbb Z$$ is not a coboundary, but I haven’t been able to do this. After all, isn’t $omega_nc_#$ just the zero map?

This is Exercise 12.24 in Rotman, and includes the hint that $pi_1(mathbb RP^n,*)congmathbb Z/2mathbb Z$ and $mathbb RP^1approx S^1$. But I didn’t use either one.