Let $G$ be a group with $y\in{G}$ and $n,r\in\mathbb{N}$. If $o(y)=n$, what is $o(y^r)$?

My attempt: Let $$o(y^r)=a,$$ Then we have $$1\_G=(y^r)^a=(y)^{ra}.$$ So we have that $$n\mid ra,$$ So either $r$ or $a$ (or both) is a multiple of $n$. I'm not too sure where to go from here or if this is even the most effective approach.
['https://math.stackexchange.com/questions/3264251', 'https://math.stackexchange.com', 'https://math.stackexchange.com/users/514050/']
Consider $g(x)=-e^{-x}$. That increases to $0$. 

Now take $f(x)=e^{-x}$. That decreases to $0$.

Clearly $$f(x)>g(x)$$

Thus, knowing that $f$ is bigger than an increasing function $g$ doesn't even prove that $f$ is increasing.
If $f$ increases more quickly than $g$, that simply means the function $f-g$ is increasing. So you want to show that $(f-g)'>0$

Your question therefore boils down to, is:

$$f-g>0\implies (f-g)'>0$$
true?

You can define $h(x)=f(x)-g(x)$ to give the statement:
$$h(x)>0\implies h'(x)>0$$
f0r which I supply $h:\Bbb R^+ \mapsto \Bbb R: h(x)=\frac1x$ as an adequate counterexample.