# $n + 1 = 2^rm$ with $m$ odd $\implies$ do not exist $2^r$ vector fields on $\mathbb{P}^n$ that are everywhere linearly independent?

What is the easiest/quickest way to see the following?

If $n + 1 = 2^rm$ with $m$ odd, then there do not exist $2^r$ vector fields on the projective space $\mathbb{P}^n$ which are everywhere linearly independent.

Thanks!

The standard way to treat statements like this one is to analyze the Stiefel-Whitney classes of $\mathbb P^n$. The full Stiefel-Whitney class of the tangent bundle is $$w(T\mathbb P^n)=(1+u)^{n+1},$$ where $u\in H^1(\mathbb P^n,\mathbb Z_2)$ is the generator.
In your particualr case $(n+1)=2^rm$ and let $p=2^r(m-1)$. Then $w_p(T\mathbb P^n)=\binom{n+1}{p}u^p=u^p\neq 0$. Here we used the fact $\binom{2^rm}{2^r}=1 \pmod 2$.
On the other hand, if there are $k$ linearly independent vector fields on $\mathbb P^n$, then $T\mathbb P^n=E\oplus \mathbb R^k$ and, therefore, $w_q(T\mathbb P^n)=w_q(E)=0$ for $q>n-k$. Combining with the fact $w_p(T\mathbb P^n)\neq 0$, we conclude $k<n-(p-1)=2^r$.