If $G$ is a finite group, then show that every element in the group has a finite order which is at most equal to $|G|$.
My attempt: Let $|G|=m$ and let $a\in{G}$ so $\text{Order}(a)=\min\{n\in{N}\mid a^n=e\}$. Now let's consider the set $S=\{a^0,a^1,a^2...a^{|G|}\}$ and since $S$ contains $m+1$ elements that all $\in{G}$ (by law of composition $a*a$) so there must be two elements which are equal i.e $a^k=a^l$. Hence all elements in $G$ have finite order at most equal to $m$.
Is this a valid proof? Thanks
Your definition of order of an element doesn't even make sense. The order of $a$ is the smallest $k$ such that $a^k=e$. And, since you used a definition which doesn't make sense, your proof doesn't make sense.
After the edition, the proof is still not valid because you were supposed to prove that $a^k=e$ for some natural $k$, but you didn't.