Let $H \triangleleft G$ such that $[G:H]=n$. Show that $g^n \in H$ for all $g \in G$. 
Let $H \triangleleft G$ such that $[G:H]=n$. Show that $g^n \in H$ for all $g \in G$.

So I've been trying this for a while and got to nowhere. So I tried to solve it by arriving to a contradiction. So take $g \in G$ such that $g^n \notin H$. Then if  $G=\bigcup_{i=0}^{n-1}Hg^i$,  I was able to arrive to a contradiction. However I wasn't able to prove that $Hg^j \cap Hg^k = \emptyset$ for every $0 \leq k,j \leq n-1$ as to be able to write $G$ as that disjoint union of coclases.
Am I on the right track? Is that true? Any hint if not?
Edit:

The idea is not to use the quotient group. It's an exercise of a chapter of Rotman's Introduction to Theory of Groups which is previous to the introduction to the quotient group.

 A: Hint: apply Langrange's Theorem (every element divides the order of the group) to the quotient group $G/H$ which has order $n$.
A: I am not sure it makes sense to do so but here is an answer which doesn't explicitly use a quotient. Note that I am in some sense proving Lagrange's Theorem in the quotient.
Lemma Let $H$ be normal in $G$.
For any $g_1,g_2,g \in G$ either $$\{g_1gH,g_1g^2H,...,g_1g^{|G|}H\} \cap \{g_2gH,g_2g^2H,...,g_2g^{|G|}H\} =\emptyset$$ or $$\{g_1gH,g_1g^2H,...,g_1g^{|G|}H\} = \{g_2gH,g_2g^2H,...,g_2g^{|G|}H\}.$$
Proof: Suppose that $g_1g^jH = g_2g^kH$, then since $H$ is normal for any $m$,  $g_1g^{j+m}H=g_1g^jHg^m=g_2g^kHg^m=g_2g^{k+m}H$ which proves the claim.
Lemma Let $H$ be normal in $G$.
For any $g_1,g_2,g \in G$ the sets $$H_1=\{g_1gH,g_1g^2H,...,g_1g^{|G|}H\}$$ and $$H_2= \{g_2gH,g_2g^2H,...,g_2g^{|G|}H\}$$
have the same number of elements.
Proof: The map from $H_1$ to $H_2$ defined by $S \mapsto g_2g_1^{-1}S$ is a bijection.
Lemma Let $H$ be normal in $G$. There exists $g_1,g_2,...,g_m$ in $G$ such that
$$\{g''H\,|\,g'' \in G\} = \bigcup_{i= 1}^{m} \{g_igH,g_ig^2H, ... , g_ig^{|G|}H\}$$
and each pair of sets in the union are disjoint.
Proof Firstly if $g' = g''g^{-1}$, then $g''H \in \{g'gH,g'g^2H, ... , g'g^{|G|}H\}$ this proves that by choosing enough $g_i$'s we will obtain the whole of $\{g''H\,|\,g \in G\}$. The first Lemma proves we can choose them all to be disjoint.
Let us assume $n < |G|$ (since for $n=|G|$ we can use Lagrange's Theorem as mentioned in a comment). Via the last two lemmas we have that $n = m |\{gH,g^2H,...,g^{|G|}H\}|$. This means that $g^kH = g^jH$ for some $1 \leq j < k < |G|$ otherwise $|\{gH,g^2H,...,g^{|G|}H\}|=|G|$. Therefore $g^{k-j} H =H$. Now suppose that $r$ is the smallest natural number such that $g^rH=H$. Then $\{gH,g^2H,...,g^{|G|}H\}=\{gH,g^2H,...,g^rH\}$. Since $g^{j}H$ and $g^{k}H$ can't be equal for $1 \leq j < k  \leq r$ otherwise $g^{k-j}H =H$ with $k-j < r$ it follows that $r$ divides $n$ and hence $g^n = (g^r)^{n/r}$ is a power of an element in $H$ and hence is in $H$.
A: We have  $\forall g \in G, Hg \in G/H$. Since $|G/H|=n$, we get 
$$\forall g \in G, (Hg)^n=H,$$
But $(Hg)^n=Hg^n$. So $\forall g \in G, Hg^n=H$, i.e. $\forall g \in G, g^n \in H$.
