Isomorphism from $Z[i]/(p)$ to $(p^n)/(p^{n+1})$ where $p$ is a irreducible element . 
Let $p\in\Bbb{Z}[i]$ be an irreducible element. Prove that for any positive integer $n\ge0$ the ideal $(p^{n+1})$ is an ideal in $(p^{n})$, and prove that multiplication by $p^n$ induces a isomorphism (defined above) between $\Bbb{Z}[i]/(p)$ and $(p^n)/(p^{n+1})$ as additive abelian groups

What I thought , was that $\Bbb{Z}[i]/(p)$ will be isomorphic to  a field with $p^2$ elements.So if I define an isomorphism 
$$\phi:\ \Bbb{Z}[i]\ \longrightarrow\ \Bbb{Z}[i]/(p),$$ 
where the kernel is the ideal $(p)$ and the other elements are going to be $(c+id )+(p)$ where $0\le c,d \le (p-1)$ .
Now $(p^n)= p^n\Bbb{Z}[i]$ and $(p^{n+1}) = p^{n+1}\Bbb{Z}[i]$. Now $(p^{n+1}) $ is going to be a subring of $(p^n)$ and if we define an isomorphism $\sigma:(p^n)-> (p^n)/(p^{n+1})$.Here the kernel will be $(p^n)$ and the other elements will be $(a+ib)p^n$ where $0\le a,b \le (p-1)$.I will probably need to show that this mapping is isomorphic to a field of size $p^2$
So this is what I have tried to do precisely.This is a question from Dummit and Foote (pg-293 Q-7(a)) .What I did not understand is what does multiplication by $p^n$ mean . Instead of writing a complete answer I would recommend some hints .

The map defined by 
$\sigma:x-> p^nx$ is an isomorphic map from $Z[i]/(p)->(p^n)/(p^{n+1})$.
We first try to see whether the map is homomorphic or not 
$\sigma((a+ib + c+id)) = p^{n}(a+c) + p^{n}(b+d)i = \sigma(a+ib) + \sigma(c+id)$.
1)injectivity:
    $\sigma(x)=\sigma(y).p^n(x)= p^n(y) .x=y$
2) surjectivity:
In $ (p^n)/(p^{n+1})$ the elements will be of the form $ p^n +(x+iy)$ where $0\le x,y\le N(p)-1$, which will be surjective as the cosets in $Z[i]/(p)$ will be of the form $ (a+ib) $ where $0 \le a,b \le N(p)-1$.
