Let $m\in\Bbb N$ be the exponent of a finite group $G\ $ ($|G|=n$). It' the smallest integer such that $g^m=e_G\ \forall g\in G$
Proving that $m\mid |G|$ is the same as proving that the least common multiple of the orders of the elements of $G$ divides the order of the group once you realize the exponent is the LCM of $o(g)$ for $g\in G$.
But how to conclude the proof?
I tried to reason with $q_i\cdot o(g_i)=|G|\ \forall i\in[n]$ but it doesn't seem to lead me enywhere
Some hints: By Lagrange's Theorem, the order of every element of a finite group must divide the order of the group. In addition it implies that $a^{|G|}=e$ for all $a\in G$.