Here's my proof, please let me know if you like it or not.
Given G is a group, and $H \trianglelefteq G$, and also $|G:H|=n$, we need to prove that for any $g\in G$, $g^n \in H$.
Proof:
Define a homomorphism $\phi: G\to G/H$, then $\ker(\phi)=H$ since $H$ is normal. $|im(\phi)|=|G/H| = |G:H| = n$ (by Lagrange). ($\phi$ is surjective because for $gh_1g^{-1} = h_2$ we have $g=h_2gh_1^{-1} \implies g \in gH$ for any $gH\in G/H$). Also, $\phi(g^n) =\phi(g)^n=(gH)^n = g^nH=H$ because for any $gH$ of order $m$, since $m \lvert n$ (by Lagrange), $(aH)^n=(aH)^{km}=((aH)^m)^k = H^k = H$.
I'm concerned a little bit about my proof in terms of my homomorphism and the overall approach. I could also have some erroneous peculiarities as well. I would appreciate your input.