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.
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No, take $\color{blue}{f(x)=1}$ and $\color{red}{g(x)=1-{1\over 2^x}}$

[![enter image description here](https://i.stack.imgur.com/VfsY1.png)](https://i.stack.imgur.com/VfsY1.png)

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If $f'(x)-g'(x)>0$ then this would be true.
Consider $f(x) = x+17$ and $g(x) = x.$ They increase at the same rate everywhere for any reasonable definition, yet $f > g.$