Prove that, when $N$ is a normal subgroup, $(gN)^a=(g^a)N$ for all $a \in \mathbb{Z}$. 
Prove that, when $N$ is a normal subgroup, $(gN)^a=(g^a)N$ for all $a\in \mathbb{Z}$.

My attempt:
I know that if $b$ belongs to $H$ then
$$bH= H\tag{1}.$$
Also, $(ge)^a$ belongs to $(gN)^a$; therefore $(g^a)N=(gN)^a$ by $(1)$.
Is my attempt correct? Can someone suggest the intuition behind this?

This question is most naturally answered in the quotient group $G/N$, whose elements are of the form $gN$, for $g\in G$. This is allowed since $N$ is normal, and thus "quotientable".
Suppose $N$ is a normal subgroup of a group $G$. The statement of the question is trivially true if $g\in N$, since in $G/N$, $gN=N\iff g\in N$.
If $g\not\in N$, then by the definition of products in $G/N$, 
$$(gN)^2=(gN)(gN)=g^2N$$
From there, you can use induction on the exponent to prove what you asked for.
