Doubt about proof of For every real $x > 0$ and every integer $n > 0$ there is one and only one positive real $y$ such that $y^n = x$ This is the Theorem 1.21 of Walter Rudin Principles of Mathematical Analysis. I follow the same reasoning of Walter Rudin except that I did take another path. Then I would like to know if my proof is ok like this. I will answer my own question with the proof. However, I used the same reasoning that Walter Rudin, but, I wonder if there are a proof more intuitive than this, because, his proof seems very tricky and mine too (In the contradiction part). Thanks in advance

Proof
Uniqueness
Since $y_1 > y_2 > 0 \implies y_1^n > y_2^n > 0$ then if $y_1 \neq y_2 \implies y_1^n \neq y_2^n$ we proved that if there exists a $y$ such that $y^n = x$ it has to be unique
Existence
Let $B = \{ t \in \mathbb{R} : x < t^n \}$
B is bounded below, since $t > 0$, $0$ is a below bound. B is nonempty, because, by Archimedian property there exists $my> x$ with $m \in \mathbb{N}, y \in \mathbb{R}, x \in \mathbb{R} $ If we put $y = 1$ we get $m > x$ if $x \geq 1$ then $m^n > x$ if $0<x < 1$ we can find that $1^n > x$ then we proved that  is nonempty, since at least $M \in B$
By Completness property, B have a infimun, let's denote it by $z = \inf B$
To prove that $z^n = x$ we wil procede by contradition
Suppose that $z^n > x$
Note that if $0 < a < b$
$b^n - a^n = (b -a)(b^{n-1} + b^{n-2}a + ... + a^{n-1})$ as $a < b$ then
$ b^n - a^n < (b-a)(b^{n-1} + b^{n-2}b + ... b^{n-1})$
$ \implies b^n - a^n < (b-a)(nb^{n-1})$ $(*)$
Let $0 < h < \frac{z^n - x}{n(z+1)^{n-1}}$
$\implies hn(z+1)^{n-1} < z^n - x $ $(**)$
Since $n(z+1)^{n-1} \neq 0$
Using $(*)$ with $a = z - h, b = z$
$z^n - (z -h)^n < hnz^{n-1}$ 
With this and $(**)$
$z^n - (z - h)^n < hnz^{n-1} < hn(z+1)^{n-1} < z^n - x$
By transition $z^n - (z - h)^n <  z^n - x$
$\implies x < (z - h)^n$ this means that $(z-h) \in B$ but $z - h < z$ and $ z$ is the infimum of $B$. This lead us to a contradiction.
Suppose $z^n < x$
same reasoning
Let $0 < h < 1$ and $h < \frac{x - z^n}{n(z+1)^{n-1}}$ $(***)$
And using $(*)$ with $a = z, b = (z +h)$
$(z + h)^n - z^n < hn(z+h)^{n-1}$
Combining this with $(***)$
$(z + h)^n - z^n < hn(z+h)^{n-1} < hn(z +1)^{n-1} < x - z^n$
$\implies (z + h)^n < x$ that means that $(z + h) \notin B$ but $z + h > z$ and since $B$ isn't bounded above, this lead us to a contradicion
Hence $z^n = x$
