Let $H$ be a normal subgroup of index $n$ in a group $G$. Show that for all $g \in G, g^n \in H$

The index of $H$ in $G$ is the order of the quotient group $G/H$. Since $[G:H] = |G/H| = n$, by Lagrange, every element of $G/H$ has order dividing $n$. Hence, every coset $gH \in G/H$ has order dividing $n$, i.e. $(gH)^{n} = g^{n}H = eH =H$, the identity coset. Thus, $g^{n} \in H$, since if $aH=bH$, then $a^{-1}b \in H$ (here $a=e, b=g^{n}$).

The multiplication in the quotient group is defined this way: $$(aH)(bH) = (ab)H$$ no need to multiply $H$ by it self.

I'm not sure what you mean by the identity, but if you're asking why $H^n=H$, then the reason is because $H^n = \{ h_1\cdot\ldots\cdot h_n \,|\, h_i\in H \, \forall i\} $, and since $H$ is a subgroup, we get that $e\in H$ ( this gives us $H \subset H^n$ ) and that $H$ is closed under the group operation ( this gives us $H^n \subset H$ ).