Integer Power Function is Bijective iff Index is Odd

Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $f_n: \R \to \R$ be the real function defined as:
 * $f_n \left({x}\right) = x^n$

Then $f_n$ is a bijection iff $n$ is odd.

Even Index
Suppose $n$ is even.

Let $x \ne 0$.

Then from Even Powers are Positive, $x^n = \left({-1}\right)^n$ and so $f_n$ is not injective.

It is equally straightforward to show it is not surjective either.

So for even $n$, $f_n$ is not bijective by definition.

Odd Index
Now suppose $n$ is odd.

From the Power Rule for Derivatives, $D_x \left({x^n}\right) = n x^{n-1}$.

As $n$ is odd, $n-1$ is even.

From Derivative of Monotone Function, it follows that $f_n$ is increasing over the whole of $\R$.

The only place where $D_x \left({x^n}\right) = 0$ is at $x = 0$.

Everywhere else, $f_n$ is strictly increasing.

So $f_n$ is definitely injective throughout all of $\R \setminus \left\{{0}\right\}$.

But $x^n = 0 \implies x = 0$ and so there is only one image of $x = 0$ under $f_n$.

It follows, then, that the whole of $f_n$ is an injection.

From Existence of Root we have that:
 * $\forall x \in \R_{\ge 0}: \exists y \in \R: y^n = x$

From Sign of Odd Power we have that $\left({-x}\right)^n = - \left({x^n}\right)$ and so:
 * $\forall x \in \R_{\le 0}: \exists y \in \R: y^n = x$

Thus:
 * $\forall x \in \R: \exists y \in \R: y^n = x$

and so $f_n$ is a surjection.

So when $n$ is odd, $f_n$ is both injective and surjective, and so by definition bijective.

Hence the result.