If the index $n$ of a normal subgroup $K$ is finite, then $g^n\in K$ for each $g$ in the group. Let $K \unlhd G$ be a normal subgroup of some group $G$ and let $|G/K|=n<\infty$. I want to show that $g^n\in K$ for all $g\in G$.
Let $g\in G$, if $g\in K$, then $g^n\in K$ and we are done. If $g^n\notin K$ then consider the set of left cosets
$$
C=\{K, gK,g^2K,...,g^{n-1}K\}
$$
I want to show that these cosets are all disjoint and hence $C=K$, then I want to show that $g^nK=K$, so $g^n\in K$. 
Suppose
$$
g^lK=g^mK
$$
for some $m,l<n$, then $g^{m-l}\in K$. I am not sure how to proceed from there.

You dont need to go that far. Since you know that $|G/K|=n<\infty$ and 
$G/K =\{K, gK,g^2K,...,g^{n-1}K\}$. Then for any $gK$ in $G/K$, $(gK)^n= g^nK=K $ which answers your question
