What you are asking for is the existence of $n$-th root. This is true in $\mathbb{R}$ but not in $\mathbb{Q}$ (for example $\not \exists q\in \mathbb{Q}:q^2=2$). Any proof of this will require the completeness of the real line (Least Upper Bound Property).
LUB property: Every subset of $\mathbb{R}$ bounded above must have a supremum.
Here is a typical proof of this theorem only using the LUB property and the Binomial Theorem.
Let $x>1$ and $S=\left\{ a>0: a^n
Since $x>1\Rightarrow x^n>x$ we have that $S$ is bounded above by $x$. Therefore, by the Least Upper Bound Property, $\exists \sup S=r\in \mathbb{R}$.
We shall prove that $r^n=x$.
Suppose that $r^n>x$ and let
\begin{equation}\epsilon =\dfrac{1}{2}\min \left\{ 1,\frac{r^n-x}{\sum\limits_{k=1}^n{\binom{n}{k}r^{n-k}}} \right\}\end{equation}
Then $0<\epsilon <1$ and so $\epsilon ^n<1$.
Therefore, by the Binomial Theorem,
\begin{gather}(r-\epsilon )^n=\sum\limits_{k=0}^{n}{\dbinom{n}{k}r^{n-k}(-\epsilon )^k}=r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}(-1)^{k-1}\ge \\
r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\epsilon ^{k-1}> r^n-\epsilon \sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}\\
(r-\epsilon)^n>
r^n-\dfrac{r^n-x}{\displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}} \displaystyle\sum\limits_{k=1}^n{\dbinom{n}{k}r^{n-k}}=r^n-r^n+x=x\Rightarrow \left( r-\epsilon \right)^n>x\end{gather}
which is a contradiction since $r=\sup S$ and $r-\epsilon
Suppose that $r^n
Therefore, $r^n=x$ if $x>1$
Let $01\Rightarrow \exists r>0:r^n=\frac{1}{x}\Rightarrow \exists r'=\frac{1}{r}>0:r'^n=\frac{1}{r}^n=\frac{1}{r^n}=x$.
If $x=1$ then $r=1$.
EDIT: Motivation as per request: As I said this is statement is not true if we replace $\mathbb{R}$ with $\mathbb{Q}$. What sets $\mathbb{R}$ and $\mathbb{Q}$ apart is the completeness of $\mathbb{R}$. The LUB property therefore must in some way be used. This is why we define $S$. Because the LUB is an existensial theorem, it shows the existence of a supremum but not its value, it is often used in proofs by contradiction.
So assuming $r^n>x$ we need to arrive to a contradiction. Remember $r$ is a very special number, the supremum of $S$. If we could show that $\exists m\in \mathbb{R}$ so that $mr$ and $m\in S$ then we are done. This is what we do with $m=r-\epsilon$ in the first case and with $m=r+\epsilon$ in the second.
It all boils down to finding an $\epsilon>0$ so that $r-\epsilon$ is an upper bound of $S$, that is $(r-\epsilon)^n>x$. We can make this $\epsilon$ as small as we want. I shall choose an $\epsilon$ for $n=2$.
\begin{equation}(r-\epsilon )^2=r^2-2r\epsilon+\epsilon^2>r^2-2r\epsilon-\epsilon\end{equation}
Remember we want $(r-\epsilon )^2>x$ and so it suffices
\begin{equation}r^2-2r\epsilon-\epsilon>x\Leftrightarrow \epsilon<\frac{r^2-x}{1+2r}\end{equation}