Find coefficient of $x^{20}$

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.

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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]]]

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.