Let $f :\Bbb R→ \Bbb R$ be given by $f(x) := x^{n}$ for some $n ∈ \Bbb N$. If $b$ is a positive real number, show that there exists a unique positive real number $a$ such that $a^{n} = b$.
My solution:
Let us choose an interval $[c,d]\subset \Bbb R$ such that $c,d>0$ and $d>c$. Now in this closed interval $f(x)$ is continuous since it is a product ($n$ times) of a continuous function $F(x)=x$. Hence $f(x)$ is continuous in $[c,d]$. Also, $f(d)>f(c)$. By Archimedean Property of reals, we can find $f(c)<b<f(d)$. By Intermediate Value Theorem, $\exists$ a point $a\in (c,d)$ such that $f(a)=b$. $\implies$ $a^{n}=b$. Since $a\in (c,d)$, $a>0$.
Proving Uniqueness of $a$:
Suppose $a$ is not unique. Then $\exists$ a $p\in \Bbb R$ such that $f(p)=b$ $\implies a^{n}= p^{n}$ $\implies p=\pm a$. If $p=-a$, then $p$ is negative since $a$ is positive. So, the only possible positive solution of $a^{n}=b$ is $p=a$. Hence uniqueness.
I just want to check if proof is rigorous enough and if there are any loopholes in the proof.
First show $\lim_{x\to \infty}x^n=\infty$. That is, $\forall M>0,\exists N\in\mathbb{N}$, such that $x^n>M, \forall x>N$.
Given $b>0$, let $f(x)=x^n$ note $f(0)=0$. By above argument, $\exists N\in\mathbb{N}$, such that $x^n>b, \forall x>N$, hence $f(N+1)>b$. Then consider $f$ on $[0,N+1]$, and apply intermediate value theorem.
By uniqueness, you only need to show $f$ is strictly increasing on $(0,\infty)$.