Order of group element divides order of finite group Proving this can be done as follows: consider a finite group G and elements $g_i \in G$ for some integer $i$. Now consider $\langle g_i \rangle = \{g_i^n: n\geq 0\}$, a generator. It can be proved that $\langle g_i \rangle \leq G$ and that the order of $g_i$ is equal to the order of $\langle g_i \rangle$, so $|\langle g_i \rangle| \leq |G|$.
We can now use Lagrange's theorem which states that if $H \leq G$ then $|H|$ divides $|G|$ and we're done. But...
Is there a simpler way to prove this fact?

Yes but it is almost same with the proof of Lagrange theorem. (as a pattern)
Choose $x\in G$, think the set $\{x,gx,g^2x,...,g^{n-1}x\}$. As you see this set has $n$ elements. Choose another $y$ which is not the set and create the set $\{y,gy,g^2y,...,g^{n-1}y\}$. ... You will see that $n$ divides order of $G$.
