Prove that $[ab]=[[a][b]]$ 
Let $a,b,N\in\mathbb{Z}$, where $N>0$. Prove that $[ab]=[[a][b]]$, where $[x]$ denotes the remainder of $x$ after division by $N$.

Here's my attempt:
Proof. Since $x\equiv{[x]}_c\pmod c$, we know $a\equiv{[a]} \pmod{N}$ and $b\equiv{[b]} \pmod{N}$. By Proposition 1.3.4 in the book,
$$ab\equiv{[a]}{[b]} \pmod N$$
Applying the first theorem we used to each side of the equation, we have
$${[ab]}={[{[a]}{[b]}]}$$
QED.
I think the last line in my proof is wrong. I tried combining these two statements:


*

*$ab\equiv[ab]\pmod{N}$

*${[a]}{[b]}\equiv[{[a]}{[b]}]\pmod{N}$


But what makes me hesitate is that the statements have equivalence, not equality. I'm new to number theory and do not know how to handle equivalence.
Can you offer any help? I am looking for hints, not complete answers.
Thank you. 

I wouldn't use congruences, personally, because $x \equiv x + N$ would confuse me.
The answer by user133281 isn't wrong, but I find it easier to do the following.
By the division algorithm, there are integers $k_i$ such that:
$[ab] = ab - Nk_{ab}$
$[a] = a - Nk_a$
$[b] = b - Nk_b$
where $0 \le [x] < N$
These are equalities, so you can use all the equality rules you're familiar with.
