Question about immersions of $\mathbb{R}P^n$ into $\mathbb{R}^{n+1}$ I am currently reading a paper which takes for granted the following geometric fact: if $\mathbb{R}P^n$ can be immersed in $\mathbb{R}^{n+1}$ then for some $k$, $n=2^k-1$ or $n=2^k-2$. My initial thought was that this has something to do with Stiefel-Whitney numbers, but I can't see how that would work. Thoughts?
 A: If $i:\mathbb P^n\hookrightarrow \mathbb R^{n+1}$ is an immersion, you have the relation on $\mathbb P^n$, involving the normal line bundle $N$:   $$ i^* T_{\mathbb R^{n+1}}=T_{\mathbb P^n}\oplus N                  $$
 From this you deduce for the total Stiefel-Whitney classes $$1=w(T_{\mathbb P^n})\cdot w(N)\in H^*(\mathbb P^n,\mathbb F_2)    \quad (*)$$
Since  $N$ is a line bundle, we have $w_i(N)=0$ for $i\gt 1$ and a little calculation  then shows that this is only possible if for some $k$ we have  $n=2^k-1$ or $n=2^k-2$.   
Edit Since I have some free time now, here is the "little calculation":
   Recall that $H^*(\mathbb P^n,\mathbb F_2)=\mathbb F_2[H]/\langle H^{n+1}\rangle= \mathbb F_2[h]$ and that $w(T_{\mathbb P^n})=(1+h)^{n+1}$.
Since $N$ has rank $1$, its total Stiefel-Whitney class is $w(N)=1$ or $w(N)=1+h$ , so we dichotomize: 
First case: $ w(N)=1$
Then from $(*)$ we get $w(T_{\mathbb P^n})=(1+h)^{n+1}=1$, hence (from arithmetic modulo $2$ : see below) $n+1=2^k$   
Second  case: $ w(N)=1+h$
Then from $(*)$ we get $w(T_{\mathbb P^n})=(1+h)^{n+1}\cdot (1+h)=(1+h)^{n+2}=1 $ and again from arithmetic modulo 2 we get $n+2=2^k$. 
Arithmetic modulo 2 fact: If $N=2^rs$ with $s$ odd then  $$(1+h)^{N}=(1+h^{2^r})^s=1\in \mathbb F_2[h] \iff 2^r\geq n+1$$
This follows from the binomial formula so dear to Professor Moriarty.
