$f(x)=x^n,n\in\mathbb{N}$, exists unique $a$ s.t. $a^n=b$ 
Let $(\mathbb{R},d_2)$ be a metric space, and let $f:\mathbb{R}\rightarrow\mathbb{R}$ be given by $f(x)=x^n$ for $n\in\mathbb{N}$. If $b\in\mathbb{R_+}$, show that there exists a unique $a\in\mathbb{R_+}$ such that $a^n=b$. 

Attempt at a proof:
By the Intermediate Value Theorem, since $\mathbb{R}$ is connected $\implies I=f(\mathbb{R})$ is an interval in $\mathbb{R}$. If $b\in f(\mathbb{R})$, $\exists a\in\mathbb{R}:f(a)=b$. So IVT implies existence, now I have to show uniqueness. Let $a,a'\in\mathbb{R_+}$: $f(a)=f(a')=(a')^n=b$. This is where I run into some trouble. I need to show that my assumptions above result in the fact that $a=a'$. 

Note that the function $x^n$ is strictly monotone on $\mathbb{R_{+}}$ particularly is injective.
(If $0\leq x<y$ then $0\leq x^n<y^n$, you show that by induction on $n$).
