Why does $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$ using ket notation? Nielsen and Chuang define $A^{\dagger}$ as the unique linear operator on $V$ such that for all vectors $|v\rangle, |w\rangle$ in $V$, $$(|v\rangle, A|w\rangle) = (A^{\dagger}|v\rangle, |w\rangle).$$
I'm having trouble deducing why $(AB)^{\dagger} = B^{\dagger}A^{\dagger}$ using only this definition.
I apply the definition to get $(|v\rangle, AB|w\rangle) = ((AB)^+|v\rangle, |w\rangle)$, but I thinking there's some other property that I can't think of, or am unaware of to prove this from scratch.
 A: Let's just do the process in two steps. Each time we just apply the rule
$$(|v\rangle, A|w\rangle) = (A^{\dagger}|v\rangle, |w\rangle) . $$
We want to apply this to $(|v\rangle, AB|w\rangle)$. Let's start by substituting $B|w\rangle\rightarrow |u\rangle$. So then we have an expression for which we can apply the rule:
$$ (|v\rangle, A|u\rangle) = (A^{\dagger}|v\rangle, |u\rangle). $$
Now we replace $|w\rangle$ and define $A^{\dagger}|v\rangle\rightarrow |t\rangle$. So now we have another expression for which we can apply the rule
$$ (|t\rangle, B|w\rangle) = (B^{\dagger}|t\rangle, |w\rangle). $$
Finally we replace $|v\rangle$. So, starting form the original expression, we get the result 
$$ (|v\rangle, AB|w\rangle) = (B^{\dagger}A^{\dagger}|v\rangle, |w\rangle), $$
which shows that the order of the operators changed. We didn't really need to do the substitutions, but it makes it obvious how to apply the rule.
Addition:
Why does 
$$ \tag{1} (A|v\rangle)^{\dagger}=\langle v|A^{\dagger} , $$
if we use $(|v\rangle)^{\dagger}=\langle v|$ as given?
There are probably various ways to show this. The way I can think of now is to say that it follows from
$$ (\langle w|A|v\rangle)^{\dagger}=\langle v|A^{\dagger}|w\rangle . $$
To see that the latter should be true one can use the trace say that
$$ \langle w|v\rangle={\rm tr}\{|v\rangle\langle w|\} $$
Now $|w\rangle\langle v|$ acts like an operator. So we can use the rule about operators to say that
$$ (A|v\rangle\langle w|)^{\dagger}=(|v\rangle\langle w|)^{\dagger}A^{\dagger} =|w\rangle\langle v|A^{\dagger} $$
This shows how the operators transform together wth the states. I could probably have gone straight to this last expression by just appending the $\langle w|$ at the end in (1).
