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. 

It is easy to see, using elementary algebra, that for
$p \in \Bbb N, \tag 0$
we have
$y^p - z^p = (y - z)(y^{p - 1} + y^{p - 2}z + y^{p - 3}z^2 + \ldots + yz^{p - 2} + z^{p - 1})$
$= (y - z)\displaystyle \sum_1^p y^{p - k}z^{k - 1}; \tag 1$
since
$y, z > 0, \tag 2$
we also have
$y^{p - 1} + y^{p - 2}z + y^{p - 3}z^2 + \ldots + yz^{p - 2} + z^{p - 1}$
$= \displaystyle \sum_1^p y^{p - k}z^{k - 1} > 0; \tag 3$
note that every term in these sums is positive.
Then if
$y^p = z^p, \tag 4$
$y^p - z^p = 0, \tag 5$
and so
$(y - z)(y^{p - 1} + y^{p - 2}z + y^{p - 3}z^2 + \ldots + yz^{p - 2} + z^{p - 1})$
$= (y - z)\displaystyle \sum_1^p y^{p - k}z^{k - 1} = 0, \tag 6$
and in light of (3) this forces
$y - z = 0 \Longrightarrow y = z. \tag 7$
$OE\Delta$.
