Find coefficient of $x^{20}$ Find the coefficient of $x^{70}$ in the expansion 

$$(x-1)(x^2-2)(x^3-3)(x^4-4)\cdots (x^{12}-12)$$

$\mathcal {\text {Now I have solved this question}}$. What I did was I noticed that the highest power that can be formed in this expansion  would be $78$ . Hence I need to find the sum of Distinct numbers that on addition would form $8$ and I have to multiply them among themselves .
Hence the answer would be simply 
$$(-8) + [(-1)(-7) + (-2)(-6) + (-3)(-5)] + [(-1)(-3)(-4) + (-1)(-2)(-5)]  = 4$$

$\mathcal {\text {But what if something like coefficient of $x^{20}$ would have been asked}}$.

Because the way I used would expend a lot of time to solve it.  I want to know if there are any standard methods to solve it. 
 A: Dividing by $x^{78}$ and substituting $x\mapsto\frac1x$ shows that
$$
\left[x^{70}\right]\prod_{k=1}^{12}\left(x^k-k\right)=\left[x^8\right]\prod_{k=1}^{12}\left(1-kx^k\right)
$$
and
$$
\left[x^{20}\right]\prod_{k=1}^{12}\left(x^k-k\right)=\left[x^{58}\right]\prod_{k=1}^{12}\left(1-kx^k\right)
$$
It is not too difficult to derive the recursion
$$
\overbrace{\left[x^m\right]\prod_{k=1}^{n\vphantom{-1}}\left(1-kx^k\right)}^{a(n,m)}
=\overbrace{\left[x^m\right]\prod_{k=1}^{n-1}\left(1-kx^k\right)}^{a(n-1,m)}
-n\,\overbrace{\left[x^{m-n}\right]\prod_{k=1}^{n-1}\left(1-kx^k\right)}^{a(n-1,m-n)}
$$
So if we define
$$
a(n,m)=\left\{\begin{array}{}
0&\text{if $m\lt0$ or $m\gt\frac{n(n+1)}2$}\\
1&\text{if $m=0$}\\
a(n-1,m)-n\,a(n-1,m-n)&\text{otherwise}
\end{array}\right.
$$
then
$$
a(12,8)=4
$$
and
$$
a(12,58)=1152600
$$
This requires less work than computing $\prod\limits_{k=1}^{12}\left(x^k-k\right)$, but it still is not simple. 
Although, when doing this by hand, the table of $a(n,m)$ gets big, it is no more difficult than computing Pascal's Triangle. For example, computing $a(12,58)$ required only $75$ values to be computed; that is less than computing $12$ rows of Pascal's Triangle.

Mathematica Code
a[n_,m_]:=If[m==0,1,If[m<0||m>n(n+1)/2,0,a[n,m]=a[n-1,m]-n a[n-1,m-n]]]
A: Let's formalize you process, which is valid.  Your polynomial is
$$
\prod_{k=1}^{12} (x^k -k)
$$
Now define $\mathbb B[12]$ the space of boolean vectors of length $12$.  (that is the set sequences of $0$ or $1$ of length $12$).
The expansion is
$$
\sum_{b\in\mathbb B[12]} \prod_{k=1}^{12} (b_k x^k) + (1-b_k)(-k)
$$
You now want to select the terms such that
$$
\prod_{k=1}^{12} (b_k x^k) + (1-b_k)(-k) = c x^{20}
$$
for some $c \in \mathbb R$.  Note that if $b_k = 1$, then $(1-b_k)(-k)$ is $0$.  So let's ignore that term for now.  You're looking for the $b \in \mathbb B[12]$ such that
$$
\prod_{\substack{k\in\{1..12\}\\b_k \ne 0}} b_k x^k = x^{20}.
$$
Which is
$$
\prod_{k=1}^{12} x^{b_k \cdot k} = x^{20}.
$$
Since $x^{a} \cdot x^{b} = x^{a+b}$, you can translate that equation on the coefficients.
$$
\sum_{k=1}^{12} b_k k = 20
$$
This is a rather recursive problem.  The number of solutions $b \in \mathbb B[12]$ is the number of solutions $b \in \mathbb B[11]$ that sums to $20$ (if $b_{12} = 0$) plus the number of solutions $b \in \mathbb B[11]$ that sums to $20-12 = 8$ (if $b_{12} = 1$).
\begin{align}
\left\{b \in \mathbb B[12]: \sum_{k=1}^{12} b_k k = 20 \right\}
&= \left\{[b,0] : b \in \mathbb B[11] \text{ and } \sum_{k=1}^{11} b_k k = 20 \right\} \\
&\cup \left\{[b,1] : b \in \mathbb B[11] \text{ and } \sum_{k=1}^{11} b_k k = 8 \right\}
\end{align}
I'll let you search a solution for the recursion if it exists (I don't have the time sorry, I feel the solution cannot be simply written anyway).
Note that for $b \in \mathbb B[12]$ that is a solution, its associated coefficient is
$$\prod_{\substack{k\in\{1..12\}\\b_k \ne 1}} (-k).$$
Hence sum for each $b \in \mathbb B[12]$ that is a solution to the above recursion, their associated coefficients to your final answer.
A: We have that
$$
P(x,n) = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {x^{\,k}  - k} \right)} 
  = \left. {{\partial  \over {\partial y}}\prod\limits_{1\, \le \,k\, \le \,n} {\left( {x^{\,k}  - y^{\,k} } \right)\;} } \right|_{\,y\, = \,1} 
$$
and
$$
\eqalign{
  & P(x,y,n) = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {x^{\,k}  - y^{\,k} } \right)\;}
  = x^{\,n\left( {n + 1} \right)/2} \prod\limits_{1\, \le \,k\, \le \,n} {\left( {1 - \left( {{y \over x}} \right)^{\,k} } \right)\;}  =   \cr 
  &  = x^{\,n\left( {n + 1} \right)/2} \prod\limits_{0\, \le \,k\, \le \,n - 1} {\left( {1 - {y \over x}\left( {{y \over x}} \right)^{\,k} } \right)\;}
  = x^{\,n\left( {n + 1} \right)/2} \left( {{y \over x};\;{y \over x}} \right)_{\,n}  \cr} 
$$
where $\left( {a;\;z} \right)_{\,q} $ denotes the q-Pochhammer .
It is known that, in terms of q-Binomial coefficents $\left( \matrix{   m \cr    k \cr}  \right)_{\,q} $
the series expansion of the $P(x,y,n)$ can be written as
$$
\eqalign{
  & P(x,y,n) = x^{\,n\left( {n + 1} \right)/2} \left( {{y \over x};\;{y \over x}} \right)_{\,n}  =   \cr 
  &  = x^{\,\left( {\scriptstyle n + 1 \atop 
  \scriptstyle 2}  \right)} \sum\limits_{0\, \le \,j\, \le \,n} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)_{\,y/x} \left( {{y \over x}} \right)^{\,j} \left( {{y \over x}} \right)^{\,\left( {\scriptstyle n \atop 
  \scriptstyle 2}  \right)} }   \cr 
  &  = \sum\limits_{0\, \le \,j\, \le \,n} {\left( { - 1} \right)^{\,j} \left( \matrix{
  n \cr 
  j \cr}  \right)_{\,y/x} y^{\,\left( {\scriptstyle n \atop 
  \scriptstyle 2}  \right) + j} x^{\,n - j} }  \cr} 
$$
But the derivative of the q-Binomial is not easily expressible.
An alternative way is to rewrite $P(x,y,n)$ as 
$$
\eqalign{
  & P(x,y,n) = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {x^{\,k}  - y^{\,k} } \right)\;}
  = y^{\,n\left( {n + 1} \right)/2} \prod\limits_{1\, \le \,k\, \le \,n} {\left( {\left( {{x \over y}} \right)^{\,k}  - 1} \right)\;}  =   \cr 
  &  = y^{\,n\left( {n + 1} \right)/2} Q(x/y,n) \cr} 
$$
and in turn, express $Q(z,n)$ as product of roots of unity
$$
\eqalign{
  & Q(z,n) = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {z^{\,k}  - 1} \right)\;}
  = \prod\limits_{0\, \le \,k\, \le \,n - 1} {\prod\limits_{0\, \le \,j\, \le \,k - 1} {\left( {z - e^{\,i\;\,\left( {j/\left( {k + 1} \right)} \right)\,2\pi } } \right)} \;}  =   \cr 
  &  = \prod\limits_{0\, \le \,j\, \le \,k\, \le \,n - 1} {\left( {z - \omega _{\,n,\,j,\,k} } \right)\;}
  = \prod\limits_{0\, \le \,l\, \le \,h - 1} {\left( {\,z - s_{\,l} } \right)}  = \sum\limits_{0\, \le \,q\, \le \,h} {c_{\,h,\,h - q} \,z^{\,q} } \quad \left| {\;h
  = \left( \matrix{ {n+1} \cr   2 \cr}  \right)} \right. \cr} 
$$
The coefficients $c_{\,h,\,h - q}$ are given by the OEIS sequence A231599.
There is a recurrence indicated therein, that is (putting the indices in bracket for better readability)
$$c(n,k) = c(n-1, k) - c(n-1, n*(n-1)/2-k)*(-1)^n \quad |\; n > 1$$
Another interesting recurrence is provided in this post
The $c_{\,h,\,h - q}$ coefficients could also be computed by resorting to Vieta's formulas.
But in doing that we shall be careful to note that the fraction $j/(k+1)$ when reduced will provide repeated values,
 and consequently repeated are the $s_l$.
$$
\eqalign{
  & \prod\limits_{0\, \le \,l\, \le \,h - 1} {\left( {\,z - s_{\,l} } \right)}  = \sum\limits_{0\, \le \,j\, \le \,h} {c_{\,h,\,h - q} \,z^{\,q} } \quad  \Rightarrow   \cr 
  &  \Rightarrow \quad c_{\,h,\,m}  = \left( { - 1} \right)^{\,m} \sum\limits_{\,\left\{ {k_{\,0} \,,\,k_{\,1} \,,\, \cdots \,,\,k_{\,m - 1} } \right\}\,\,
 \subset \;\left\{ {0\,,\,1\,,\, \cdots \,,\,h - 1} \right\}\;\;} {\prod\limits_{0\, \le \,j\, \le \,m - 1} {s_{\,k_{\,j} } } }  \cr} 
$$
After which , going back to $P(x,y,n)$ and derive wrt $y$ is quite straight. 
Otherwise, it is possible to start directly from $P(x,n)$, rewritten as
$$
\eqalign{
  & P(x,n) = \prod\limits_{1\, \le \,k\, \le \,n} {\left( {x^{\,k}  - k} \right)}  =   \cr 
  &  = \prod\limits_{0\, \le \,k\, \le \,n - 1} {\;\prod\limits_{0\, \le \,j\, \le \,k - 1} 
 {\left( {x - \left( {k + 1} \right)^{\,1/\left( {k + 1} \right)} e^{\,i\;\,\left( {j/\left( {k + 1} \right)} \right)\,2\pi } } \right)} \;}  \cr} 
$$
and applying Vieta's formulas as above.
Although algebraically interesting, I am however concious that this approach is not computationally viable for large $n$.
