Proving the odd integer power function is injective over the reals I am taking Abstract Algebra and in the first assignment we were asked to determine which values of $n$ make the following function injective.
$f: \mathbb{R}\rightarrow \mathbb{R}$
$x \mapsto x^n$ $| n \in \mathbb{N^+}$ 
Obviously the case where $n$ is even is quite easy to disprove. For the odd case I had a more difficult time. I understand that I could use the fact that the odd functions are continuous and because their derivative is positive everywhere (besides 0 which is dealt with separately), the function is increasing so $a > b \implies f(a) > f(b)$ which would prove it was injective. However, I do not think this is the proper way to go about it, as we have not and will not cover continuity and those sort of things. Not to mention this method seems out of place with the other problems in the homework which all deal with equivalence relations and general set theory questions.
So I just want to know, is there a simpler way of going about proving the odd power function is injective that does not use much Real Analysis as much?
 A: Take any $x,y$. If $x,y$ have different signs, so do their odd powers, so they are distinct. So we can assume that they have the same sign, wlog both are positive. Then $$x^{2n+1}-y^{2n+1}=(x-y)\left(\sum_{j=0}^{2n}x^jy^{2n-j}\right)$$
And the second parenthesis is a positive expression, so the expression is zero iff $x=y$.
A: Here's an algebraic proof. Suppose $a^{2m+1} = b^{2m+1}$. Then $(a/b)^{2m+1} = 1$, which reduces the problem to showing that $1$ is the only real number $z$ with $z^{2m+1} = 1$. The rest can be done with basic abstract algebra. First, a well known (and easy to prove) lemma on the number of roots to a polynomial over a field shows that there are at most $2m+1$ roots to $x^{2m+1} -1 = 0$ in $\mathbb{C}$. We can enumerate exactly $2m+1$ distinct $(2m+1)$st roots of unity (as complex powers of $e$) and verify that only one of them is real, namely $z = 1$.  Thus $z^{2m+1}$ is injective over the reals.
A: One way to know using elementary calculus is this. The case $x\mapsto x$ is trivially injective.   Suppose $n$ is a positive integer. Then
$$ \left(x^{2n + 1}\right)' = (2n + 1)x^{2n} > 0 \qquad {x \not= 0}.$$
By calculus, the function $x\mapsto x^{2n+1}$ is strictly increasing and is therefore 1-1.
