Isomorphism of quotient rings in algebraic number theory Let $\zeta$ be a $p$-th root of unity. I read the following sequence of isomorphisms:
$$\mathbb{Z}[\zeta]/p\mathbb{Z}[\zeta] \simeq \mathbb{Z}[X]/(p, \Phi_p) \simeq (\mathbb{Z}/p\mathbb{Z})[X]/\Phi_p \simeq (\mathbb{Z}/p\mathbb{Z})[X]/(X-1)^{p-1}$$
All this seems to be based on a certain "associativity" of the order in which we can take the quotient, I think I haven't seen this stated clearly in my algebra course. But I mainly don't get the last point: do we really have
$$\Phi_p(X) \equiv (X-1)^{p-1} \quad \text{mod } p?$$

By Eisenstein's criterion, it is easy to show that $\frac{x^p-1}{x-1}$ is irreducible for any prime $p$ (substitute $x=y+1$).  
And $\mathrm{deg}\left(\frac{x^p-1}{x-1}\right)=p-1.$
So $\Phi_p(X)=\frac{X^p-1}{X-1}$.
Now, consider $\Phi_p(X)-(X-1)^{p-1}$. This is clearly a polynomial of degree $p-1$. 
By Fermat's little theorem, for any $a\not\equiv 1\;\mathrm{mod}\; p$ we have $(a-1)^{p-1}\equiv 1\; \mathrm{mod}\; p$. 
But also, for any $a\not\equiv 1\;\mathrm{mod}\; p$ we have $\Phi_p(X)=\frac{a^p-1}{a-1}\equiv \frac{a-1}{a-1}\equiv 1\; \mathrm{mod}\; p$. 
So for any $a\not\equiv 1\;\mathrm{mod}\; p$, we have $$\Phi_p(a)-(a-1)^{p-1}\equiv 1-1= 0\; \mathrm{mod}\; p$$
If, on the other hand, $a\equiv 1\; \mathrm{mod}\; p$, then $(a-1)^{p-1}\equiv 0\; \mathrm{mod}\; p$. 
But (given that, in general, $\frac{x^n-1}{x-1}=1+x+x^2+\dots+x^{n-1}$) we have also that $$\Phi_P(1)\equiv\underbrace{1+1+\dots+1}_p=p\equiv 0\; \mathrm{mod}\; p$$ 
So, as before: $$\Phi_p(a)-(a-1)^{p-1}\equiv 0-0=0\; \mathrm{mod}\; p$$
So $\Phi_p(X)-(X-1)^{p-1}$ has all of $\{0,1,2,\dots,p-1\}$ as roots. But these are all distinct in $\mathbb F_p$ (mutually incongruent). 
Since $\mathrm{deg}\left(\Phi_p(X)-(X-1)^{p-1}\right)=p-1$ and $\Phi_p(X)-(X-1)^{p-1}$ has $p$ roots, we conclude $\Phi_p(X)-(X-1)^{p-1}\equiv 0\; \mathrm{mod}\; p$
