Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in H$ for all $g \in G$

Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in
H$ for all $g \in G$

I need some help in relation to this exercise
"Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in
H$ for all $g \in G$."
I tried by induction on $n$. The case $n=1,n=2$ are obvious, but even the
case $n=3$ is giving me trouble so I give up studying the general case of
the inductive step.
My other approach was studying left or right coset of $G$. But I only
proved that if $g \in aH$ then $g^2 \notin aH$ if $a \notin H$, and I
can't find a way to demonstrate that $g^n \in H$. (my starting idea was to
prove that every power of $g$ is in a different coset but then I realize
that in this way I don't handle several case, for example $g$ has period
strictly lesser than $n$ and in conclusion it doesn't prove the exercise)
Maybe I'm missing something about indexes, and this is why I asked here
for some help,
(I can't use quotient groups because they are introduced later than this
exercise, forgot to add this info in the beginning) Thanks in advance :)