I do not fully understand the proof for even roots so I cannot apply it for odd.
The proof for even $n$ uses Bolzano-Weierstrass theorem which I understand.
What I don't understand is the restriction of domain of $f(x) = x^n$ to $[\ 0, \infty)$. One can then show that the codomain is $[\ 0, \infty)$. Why one does that?
After that the whole proof goes by setting the grounds for using the Bolzano-Weierstrass theorem to prove that $x^n$ is surjective, continuous.
Since we know that $x^n$ for even $n$ is injective on the picked interval we have proved that function is bijective which means that its inverse is also continuous.
I was thinking of using the $x^{2n+1}$ notation for odd $n$, provide a proof for even and then, using the fact that product of continuous functions is continuous, prove that $x^{2n}x$ is continuous. I was considering this only because I don't know how to apply the proof for even roots to odd(for odd $n$ function is injective on the entire domain).
But then I got stuck on proving that $x$ is continuous.