Prove that $ y = x^{\frac{1}{n}} \Rightarrow y^n = x $ Tao, Analysis I, exercise 5.6.1

I have to prove the following where $x,y \in \Bbb R^+, \ n \in \Bbb Z^+$
$$ y = x^{\frac{1}{n}} \Rightarrow  y^n = x $$
Hints: review the proof of Proposition
  5.5.12. Also, you will find proof by contradiction a useful tool, especially when
  combined with the trichotomy of order in Proposition 5.4.7 and Proposition
  5.4.12.

My attempt:
Assume that $ y = x^{\frac{1}{n}}  \Rightarrow  y^n > x $. By definition: 
$$x^{\frac 1n}= \sup \{ y \in \mathbb R, s.t.  y \geq 0, y^n \leq x \}$$
Thus  $ y^n > x $ contradicts the above definition, since $x$ is upper bound of $y^n$.
Does this part seem ok? I next need to get to a contradiction starting from here:
Now assume that $ y = x^{\frac{1}{n}}  \Rightarrow  y^n < x $
 A: Let $a := x^{1/n}$.
Then $a^n > x$ does not conflict with the definition of the set, since it merely states that $a^n$ is an upper bound.
Here is my version of the proof following the same mechanism than for 

Proposition 5.5.12. There exists a positive real number $x$ such that
  $x^2 = 2$.

and using the result of

Exercise 5.4.4. Show that for any positive real number $x > 0$ there
  exists a positive integer $N$ such that $x > 1/N > 0.$

Assume that $a^n > x$. Then there exists $\varepsilon > 0$ such that $(a - \varepsilon)^n > x$ and $a - \varepsilon$ is not in the set $\{y \in \mathbb{R}: y \ge 0 \; \wedge \; y^n \le x\}$. This contradicts the fact that $a - \varepsilon$ is not an upper bound, since $a$ is least upper bound, by definition.
Assume that $a^n < x$. Then there exists $\varepsilon > 0$ such that $(a + \varepsilon)^n < x$, and therefore there is an element in $\{y \in \mathbb{R}: y \ge 0 \; \wedge \; y^n \le x\}$ larger than $a$. But this contradicts that $a + \varepsilon > a$ is an upper bound.
In effect, $a^n \le x$ and $a^n \ge x$, equivalent to $a^n = x$.

To see that there exists an $\varepsilon > 0$ such that $(a + \varepsilon)^n < x$, note that $a$ is bounded by $x$, if $x > 1$. And it is bounded by $1$ otherwise. This yields
$$
(a + \varepsilon)^n = \sum_{k=0}^n  
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}
\varepsilon^k
= a^n + \varepsilon
\left(
\sum_{k=1}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}
\varepsilon^{k-1}
\right)
<
\begin{cases}
a^n + \varepsilon
\left(
\sum_{k=1}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
x^{n-k}
\varepsilon^{k-1}
\right) \; \text{if} \; x > 1, \\
a^n + \varepsilon
\left(
\sum_{k=1}^n
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
\varepsilon^{k-1}
\right) \; \text{otherwise}.
\end{cases}
$$
In both cases there exists a positive real number $y$ such that $(a + \varepsilon)^n < a^n + \varepsilon y$. 
Now if $a^n < x$, we have that $x - a^n =: \delta > 0$. But for every real number $\delta$ there exists a natural number $N$ such that $\delta > N^{-1}$. Choosing $\varepsilon = y^{-1}N^{-1}$ yields $x - a^n > \varepsilon y \Rightarrow x > (a + \varepsilon)^n$.
Equivalently, to see that there exists an $\varepsilon > 0$ such that $(a - \varepsilon)^n > x$, expand $(a - \varepsilon)^n$:
$$
(a - \varepsilon)^n = a^n + (-\varepsilon)\sum_{k = 1}^n 
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}(-\varepsilon)^{k-1}
= 
a^n +
\varepsilon
\sum_{k = 1}^n
(-1)^k
\left(
\begin{array}{c}
n \\
k
\end{array}
\right)
a^{n-k}\varepsilon^{k-1}.
$$
Splitting the sum in a positive and a negative part:
$$
(a - \varepsilon)^n = a^n + \varepsilon
\left(
\sum_{i=2k}^n 
\left(
\begin{array}{c}
n \\
i
\end{array}
\right)
a^{n-i}\varepsilon^{i-1}
- \sum_{j=2k - 1}^n
\left(
\begin{array}{c}
n \\
j
\end{array}
\right)
a^{n-j}\varepsilon^{j-1}
\right),
$$
$$
(a - \varepsilon)^n
>
a^n - \varepsilon
\left(
\sum_{j=2k - 1}^n
\left(
\begin{array}{c}
n \\
j
\end{array}
\right)
a^{n-j}\varepsilon^{j-1}
\right)
$$
But, again, since $a$ is bounded by $\max(1, x)$, there is a positive real number $y$ such that $(a - \varepsilon)^n > a^n - \varepsilon y$. 
Now if $a^n > x$, we have that $a^n - x =: \delta > 0$. Following the same argument than above, $a^n - x > \varepsilon y \Rightarrow -x > \varepsilon y - a^n \Leftrightarrow x < a^n - \varepsilon y < (a - \varepsilon)^n$.
A: All numbers in this answer are assumed to be positive.
We present the following general result from real analysis:
Proposition 1: Let $f: (+0, +\infty) \to (+0, +\infty)$ satisfy the following conditions:
$\tag 1 f \text{ is strictly increasing}$
$\tag 2 (\forall \, \gamma, \text{ integer } k) (\exists \, a, b) \bigr [ a \lt b \land f(b) - f(a) \lt k^{-1} \land f(a) \lt \gamma \lt f(b) \bigr ]$
Then $f$ is a bijection (i.e. an automorphism of the total ordering structure). Moreover,
$\tag 3 f^{-1} (\gamma) = \text{sup}(\{a \mid f(a) \lt \gamma\}) = \text{inf}(\{b \mid f(b) \gt \gamma\})$
The proof is left as an exercise.
All that remains is to show that $f(x) = x^n$ satisfies $\text{(1)}$ and $\text{(2)}$.
Hint: Use the identity
$\quad b^n - a^n = (b - a) \, \displaystyle{\sum_{i=1}^n a^{n - i} b^{i - 1}}$
