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. 
 A: $$y^p=z^p\implies (y-z)(y^{p-1}+y^{p-2}z+y^{p-3}z^2+\cdots +z^{p-1})=0$$
But since both $y,z$ are positive,
$$y^{p-1}+y^{p-2}z+y^{p-3}z^2+\cdots +z^{p-1}>0$$
So $y-z=0\implies y=z$
A: 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 $f(x)=f(y)\implies x=y$.
A: 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
$y - z = 0 \Longrightarrow y = z. \tag 7$
$OE\Delta$.
A: If you know about logarithms and exponentials, you could also observe that $x^p = y^p$ implies $\ln(x^p)= \ln(y^p)$, that is, $p\ln(x) = p\ln(y)$. It follows that $\ln(x) = \ln(y)$, whence $e^{\ln(x)} = e^{\ln(y)}$, that is, $x = y$.
A: Are you familiar with induction and/or the well ordering principal?
If we assume $y \ne z$ and wolog we assume $0 < y < z$ then $y^1 < z^1$.  Let's assume there is a $y^k \ge z^k$.  By well ordering principal there most be so first $k$ where that is true.
But that would mean $y^{k-1} < z^{k-1}$.  Byt then $y^k = y^{k-1}*y < z^{k-1}*y < z^{k-1}*z = z^k$ and thus we have a contradiction.
Alternatively by induction as $y^1 < z^1$ and $y^{k-1} < z^{k-1}\implies $y^k < z^k$ we can never have anything *but* $y^p < z^p$.
So if $y^p = z^p$ we can't have $y < z$ and by wolog (or symmetry) we can't have $y > z$ either.  So if $y^p = z^p$ having $y = z$ is our only option.
....
or.  Consider $y^p = z^p$ so $y^p - z^p = 0$ and notice (although this requires induction to actually state) that $(y-z)(y^{p-1} + y^{p-2}z + y^{p-3}z^2 + .... + y^2z^{p-3} + yz^{p-2}+z^{p-1}) = y^p - z^p=0$.
Now all $y^iz^k > 0$ so $(y^{p-1} + y^{p-2}z + y^{p-3}z^2 + .... + y^2z^{p-3} + yz^{p-2}+z^{p-1}) \ne 0$.  So $(y-z)(y^{p-1} + y^{p-2}z + y^{p-3}z^2 + .... + y^2z^{p-3} + yz^{p-2}+z^{p-1}) = 0 \iff (y-z) = 0$.  So $y-z = 0$ and $y= z$>
If $0 < y < z$ then $y^1 < z^1$ so we can't have $x^p =z^p$ if $p=1$.
Now if for some $p=k$ that $y^k < z^k$ then $y^{k+1} = y^k*y < z^k*y$ and $z^k *y < z^k*z = z^{k+1}$.  So by induction it is true that $y^p < z^p$ for all $p$ and so if $y^p = z^p$ it 
