Let $N$ be a normal subgroup of index $m$ in $G$. Prove that $a^m \in N$ for all $a \in G$.

I'm trying to understand this proof:

Let $N$ be a normal subgroup of index $m$ in $G$. Prove that $a^m \in > N$ for all $a \in G$.
Proof $\;\;$ Let $a\in G$. Since $[G:N]=m$, then $|G/N|=m$. From Lagrange's Theorem, it follows that $(aN)^m=a^mN=eN=N$. Hence $a^m \in N$.

I don't understand why $a^mN=eN$. Can anybody help me out ?

$m$ is the order of the quotient group $G/N$, $aN$ is an element of that group. What happens when you raise an element to the order of a group? What is the identity element of the group $G/N$?

It follows from Lagrange's theorem, since the order of the quotient group $G/N$ is $m$. Therefore, any element of $G/N$ (all of which are of the form $aN$ for some $a\in G$) satisfies $\left(aN\right)^m = e_{G/N} = eN = N$. We also know that to compute a product $kNhN$ in a quotient group $G/N$, we can simply look at the product of the representatives $k$ and $h$ and the product $kNhN$ will be $khN$. Therefore, we have $e_{G/N} = eN = N = \left(aN\right)^m = a^m N$, so $a^m\in N$.