Proving that $x^n=a$, for $n>0$ an odd natural number, has exactly one real root In my school book, I read this theorem

Let $n>0$ is an odd natural number (or an odd positive integer), then the equation $$x^n=a$$ has exactly one real root.

But, the book doesn't provide a proof, only tells $x=\sqrt [n]a$.
How can I prove this theorem?
I tried to prove some special cases
$$x^3=8$$
$$(x-2)(x^2+2x+4)=0$$
$$x=2 \vee x^2+2x+4=0$$
But the Discriminant of $x^2+2x+4=0$ equals to $2^2-4×4=-12<0$. So $x=2$ is an only root. But for $x^5=32$, I got $x=2$ and $x^4+2x^3+4x^2+8x+16=0$.
I don't know how I can proceed.

If $n$ is even then $x^n=a$ has two real roots, $x=\pm\sqrt[n]{a}$, since $x^n=\left(x^{(n/2)}\right)^2$ is always positive, but solutions are restricted to $a\geq0$. If $n$ is odd then $x^n=a$ does not have the negative root $x=-\sqrt[n]{a}$ since $(-\sqrt[n]{a})^n=(-1)^n (\sqrt[n]{a})^n=-a\neq a$. It may also have solutions when $a$ is negative.
One can prove uniqueness of the solution to $\sqrt
[n]{a}$ using properties of the real numbers.  If there is a number such that $b^n=(\sqrt[n]{a})^n=a$ then $(\sqrt[n]{a})^{-n}b^n=1$ thus $b\sqrt[n]{a}=1$ and by uniqueness of inverse $b=\sqrt[n]{a}$. (If $n$ is even you also need to take into account negative roots.)
