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.
 A: To prove this you can do the following to show that the function $y=x^n$ is increasing when $n$ is odd.
So suppose $a\gt b$ and $n$ is odd, we want to prove $a^n\gt b^n$. Well if $a\gt 0 \gt b$ then you are adding the positive terms $a^n$ and $-b^n$. Else $$a^n-b^n=(a-b)\left(a^{n-1}+a^{n-2}b+\dots b^{n-1}\right)$$
Here $a-b$ is positive by hypothesis and every term $a^rb^{n-1-r}$ is non-negative because $n$ is odd and $a$ and $b$ do not have opposite signs (one of them could be zero). Finally since $a\gt b$ we have $a\neq b$ so that either $a^{n-1}$ or $b^{n-1}$ is positive (non zero) so $a^n-b^n$ is the product of two strictly positive numbers and is positive.
A: We can assume wlog $x$ and $a$ both positive such that $x^n=a$ indeed
$$x^n=a \iff (-x)^n =(-1)^nx^n =-a$$
Then assume by contradiction $\exists y>0 \; y\neq x$ such that $y^n=a$ then
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots xy^{n-2}+y^{n-1})=0$$
which is impossible, that is $x^n$ is injective.
Therefore it suffices to show that at least one solution exists and it follows from IVT using that $x^n$ is continuous with $x^n=0$ at $x=0$ and $\lim_{x\to \infty}x^n = \infty$, that is $x^n$ is also surjective.
A: 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.)
A: The case $a=0$ is obviously trivial. Suppose that $a>0$. This implies $x^n>0\implies x>0$, where $n$ is an odd positive integer.
Thus, we can apply the real-valued logarithm rules to the both sides:
$$
\begin{aligned}x^n=a,\,a>0
&\implies \ln x^n =\ln a\\
&\implies n \ln x= \ln a\\
&\implies \ln x =\frac {\ln a}{n}\\
&\implies x=e^{\frac {\ln a}{n}}\end{aligned}$$
Then, note that $f(x)=e^x$ is an exponential function. By properties of the real-valued exponential function $f(x)=e^x$ is strictly increasing function and the value of $e^x$ is an unique. This implies that, the value of $e^{\frac {\ln a}{n}}$ is unique.
On the other hand,
$$
\begin{aligned}x=\left(e^{\ln a}\right)^{\frac 1n}=a^{\frac 1n}=\sqrt [n]{a}.\end{aligned}
$$
Then suppose that, $a<0$. Since $n$ is an odd positive integer, we have:
$$
\begin{aligned} x^n=a\iff(-x)^n=-a>0 \end{aligned}
$$
Now applying the same logarithm rules, we get
$$-x=e^{\frac {\ln (-a)}{n}}\implies x=-e^{\frac {\ln (-a)}{n}}$$
This means, value of $-e^{\frac {\ln (-a)}{n}}$ is also unique.
On the other hand,
$$
\begin{aligned}
(-x)^n=-a,\,-a>0&\implies (-x)=\sqrt [n]{-a}\\
&\implies -x=-\sqrt [n]{a}\\
&\implies x=\sqrt[n]{a}.\end{aligned}
$$
