Prove $y = z$ if $y^p = z^p$ where $p \in \mathbb{N}$ and $y, z \in \mathbb{R}_{++}$ As above. I am beginner in analysis. I don't know how rigorously you have to prove things. To me it's obvious. 

Let $f(x)=x^p$ and $p\in\mathbb N^*$ then $f'(x)=px^{p-1}>0$ for $x>0$.
So $f \nearrow$ strictly on $(0,+\infty)$, which implies injectivity.
Thus $f(x)=f(y)\implies x=y$.
