Virasoro descendants in Big Yellow Book Equation 6.137 (page 175) of the Conformal field theory book by Philippe Di Francesco, Pierre Mathieu, David Sénéchal, is 
$$(L_{-n-2}A)(w)=\frac{1}{n!}(\partial^nTA)(w)$$.
Is this equivalent to 
$$(L_{-n-2}A)(w)=\frac{1}{n!}(L_{-1}^nL_{-2}A)(w)?$$
It seems this is not correct, because for example $L_{-1}L_{-2}A=L_{-3}A+\#L_{-2}L_{-1}A$, where the second term is not zero. And the second term $L_{-2}L_{-1}A\ne \partial L_{-2}A $.
How should I actually understand equation 6.137 of the book?
 A: You need to be careful with these manipulations. The crucial idea is that when $T(z)$ is some conserved current, the authors denote by $(TA)(w)$ the expression
$$\frac{1}{2\pi i} \int_w dz\, \frac{T(z)A(w)}{z-w}$$
where $\int_w$ is a contour integral around $z=w$, and in particular
$$(L_{-n-2}A)(w) = \frac{1}{2\pi i} \int_w dz\, \frac{T(z)A(w)}{(z-w)^{n+1}}\,.$$
This defines a local operator at the point $w$. The notation $(\partial^n T A)(w)$ is just a shorthand, which follows from the fact that
$$\frac{1}{2\pi i} \int_w dz\, \frac{T(z)}{(z-w)^{n+1}} = \frac{1}{n!} (\partial^n T)(w)$$
(a variant of Cauchy's theorem). 
Your question is really based around one false manipulation. Whereas it's true that
$$(TA)(w) = (L_{-2}A)(w)$$
(this is how we define the action of $L_{-n}$!), it's definitely not true that $T(z) \neq L_{-2}$. The former is a local operator, the latter a generator of the Virasoro algebra. 
In the future it's good to remind yourself that acting with one of the $L_{-n}$ always comes down to taking a contour integral of $T(z)$. More generally, if you have a product of generators, like
$$(L_{-n_1} L_{-n_2} \dotsm L_{-n_k} A)(w)$$
this means that you have to do $k$ contour integrals. Starting from this prescription, you recover the correct Virasoro commutator $[L_m,L_n]$. With your prescription, you would conclude that
$$L_{-n-2} = \frac{1}{n!} L_{-1}^n L_{-2}$$
which is simply wrong.
A: Let us accept that $ \partial X=L_{-1}X$ for any field $X$. Applying this to $X=A$ and $X=(TA)=L_{-2}A$, we get 
$$
(\partial TA) = \partial(TA) - (T\partial A)= [L_{-1},L_{-2}]A = L_{-3}A
$$
Iterating, we get the relation in the Big Yellow Book. In terms of the Virasoro algebra, this relation amounts to
$$
L_{-n-2} = \frac{1}{n!}\operatorname{ad}_{L_{-1}}^n (L_{-2})
$$
where $\operatorname{ad}_{L_{-1}}(Y) = [L_{-1},Y]$.
