Prove that $f(x) = x ^ {2n+1}$ is injective. So I need to prove that 
$$f : \Bbb R \rightarrow \Bbb R,\quad f(x) = x ^{2n+1},$$
where $n\in\Bbb N$ is an injective function. Or rather I need to prove that the function of a number to an odd exponent is injective.
I've been trying to prove it with a contradiction, assuming that there are 2 different values of $x$ for the same $y$ but then I get stuck.
Any help would be highly appreciated!
 A: Take $x, y$ such that $x^{2n+1} = y^{2n+1}$. If $x = 0$, the expression becomes just $0 = y^{2n+1}$, which means that $y = 0$ as well. Same for $y = 0$ implying $x = 0$.
Assuming $x, y\neq 0$, they must clearly have the same sign, because otherwise $x^{2n+1}$ and $y^{2n+1}$ wouldn't have the same sign, and couldn't possibly be equal. Finally, note that $x^{2n+1} = y^{2n+1}$ may be rewritten to $x^{2n+1} - y^{2n+1} = 0$, which gives
$$
0 = x^{2n+1}-y^{2n+1} = (x-y)(x^{2n} + x^{2n-1}y + x^{2n-2}y^2 + \cdots + xy^{2n-1} + y^{2n})
$$
Because $x$ and $y$ are both non-zero, and have the same sign, the second factor is clearly positive (it's a sum of $2n+1$ strictly positive terms). So if the product is $0$, that means that the first factor must be $0$. This gives $x-y = 0$ or $x = y$.
A: Suppose $x^{2n+1} = y^{2n+1}$. The numbers $x,y$ must have same sign. We're obviously interested in the nonzero case so $$\frac{x}{y} = \left (\frac{y}{x}\right )^{2n} $$
If $x>y$, then $1<\frac{x}{y} = \left (\frac{y}{x}\right )^{2n}<1$?!
Similarly, if $x<y$, then $1>\frac{x}{y} = \left (\frac{y}{x}\right )^{2n}>1$?!

Arthur's much better explanation: for the nonzero case
$$1 = \left (\frac{x}{y}\right )^{2n+1} \implies 1 = \frac{x}{y} $$
since the number $1$ has exactly one real $2n+1$st root.
