If $G$ is a finite group, prove that, given $ a \in G $, there is a positive integer $n$, depending on $a$, such that $a^n=\bf1$, where $\bf1$ is the identity element of the group.
This is my proof:
Since the group is finite, it's not the case that $ \forall m,n \in \mathbb{N}, a^m\neq a^n$, therefore $ \exists p,q \in \mathbb{N}$ such that $a^p=a^q$. Supposing wlog that $p>q$, $a^{p-q}=\bf1$, hence the thesis.
My doubt is about the first assumption. It is intuitive to me that equal two powers of $a$ must exist if the group is finite, but I'm afraid I haven't justified it the proper way.
You haven't quite justified it, though I doubt anyone would argue with it either. The thing to note is that there are infinitely many terms $a^m$ and all of them are in a finite group - and you certainly can't fit infinitely many distinct terms into a finite set. In fact, with the pidgeonhole principle, you can prove that there is a pair $a^m=a^n$ in the first $|G|+1$ powers of $a$.