Prove that any positive number has an $n$th root If a real number $a>0$ has an $n$th root, then it is unique: If $x^n=a=y^n$ (with $x$ and $y$ also positive), then $x=y$, because
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+xy^{n-2}+y^{n-1}),$$
and the factor on the right is positive. This also shows that $x^n<y^n$ if and only if $x<y$.
Now consider the set
$$S=\{x\mid x>0,\;x^n<a\}.$$
This is non-empty: If $a\geq1$, then $(\tfrac12)^n<a$; and if $a<1$, then $a^n<a$.
It's bounded above: If $a\geq1$, then $a^n\geq a$, so $a\not\in S$, and for any $x>a$, it follows that $x^n>a^n\geq a$, so $x\not\in S$. And if $a<1$, then $1^n>a$, so $1\not\in S$, and likewise for any $x>1$.
Therefore, $S$ has a supremum $M>0$. How can we show that $M^n=a$?
These three answers suggest that it's easy, but I don't see it.
We could define $\sqrt[n]{a}=\exp(\tfrac1n\ln(a))$, but I don't want to rely on transcendental functions to construct this simple algebraic function. And I don't want to refer to the Intermediate Value Theorem or continuity.

If $M^n<a$, then there is some number between these (for example $\frac{M^n+a}{2}$); if this could be written in the form $x^n$, then $M^n<x^n<a$ would imply $M<x\in S$ and contradict the definition of supremum. But how can we find such $x$? Maybe write $x=M+\varepsilon$ and use the binomial expansion...
Similarly, if $M^n>a$, and something between them could be written in the form $N^n$, then $N$ would be a smaller bound on $S$ than $M$, contradicting the definition of supremum. How can we find such $N$?
 A: So, given $a>0$, $M>0$, $M^n\ne a$, you want to construct $x$ such that
either $M^n<x^n\le a$ or $M^n>x^n\ge a$.
Try $x=My$. Then you want to find $y$ such that $1<y^n\le N$ or
$1>y^n\ge N$ where $N=a/M^n\ne1$.
Let's do the second case, where $N<1$ first. Let $N=1-t$, and take $y=1-t/n$.
Bernoulli's inequality implies
$$1>y^n=\left(1-\frac tn\right)^n\ge 1-t=N.$$
The first case, where $N>1$ reduces to the second. There is $y'$ with
$1>y'^n\ge 1/N$ so take $y=1/y'$.
A: One useful tool when you do not already posses all the machinery from calculus is usually Bernoulli inequality i.e. (under suitable assumptions on the domain of $x$, see the Wikipedia page):
$$(1+x)^n\ge 1+nx\\
(1-x)^n\ge 1-nx$$
In our case, the contradiction can be obtained quite easily: if $M^n<a$ then we could find a number $\delta$ such that $\frac{1}{M}>\left(\frac1{nM}\left(1-\frac{M^n}{a}\right)\right)>\delta>0$.
Considering 
$$\left(\frac{1}{M}-\delta\right)^n=\frac{1}{M^n}\left(1-\delta M\right)^n>\frac{1}{M^n}\left(1-n\delta M\right)>\frac1a$$.
Thus $M^n<\left(\frac{1}{M}-\delta\right)^{-n}=\frac{M^n}{(1-\delta M)^n}<a$, and this contradicts the assumption that $M$ is the least upper bound. 
Similarly, a contradiction is obtained considering $M^n>a$ by setting $0<\delta<\frac{M}{n}\left(1-\frac{a}{M^n}\right)$
A: Rudin does this well. The identity you gave implies $$b^n -a^n <(b-a)n b^{n-1}$$ whenever $0<a<b$. Then assume $$M^n <x.$$ Let $h$ be such that $0<h<1$ and 
$$h< \frac{x-M^n}{n(M+1)^{n-1}}.$$ Ok, seems random, but then $$(M+h)^n -M^n < hn(M+h)^{n-1} <hn(M+1)^{n-1} <x-M^n.$$Therefore, $$M+h \in S.$$ But since $M+h>M$, we are contradicting the fact that $S$ is bounded above by $M$. So $$x\leq M^n.$$ Now assume $M^n >x$. Let $$k = \frac{M^n -x}{nM^{n-1}}.$$ Clearly $k<M$. If $r\geq M-k$ then $$M^n -r^n \leq M^n -(M-k)^n <knM^{n-1} =M^n-x.$$ Therefore, $r^n >x$ and $r\notin S$. Thus, $M-k$ is an upper bound of $S$. But wait a second, we claimed $M$ to be the least upper bound of $S$, not $M-k$. Hence, $$M^n=x.$$
