<s>[INST] Find the coefficient of $x^{70}$ in $(x^1-1)(x^2-2)(x^3-3)\cdots(x^{12}-12)$. 
Find the coefficient of $x^{70}$ in $(x^1-1)(x^2-2)(x^3-3)\cdots (x^{12}-12)$.

I tried to solve this problem using theory of equation the coefficient of $x^{70}$ will be the  sum of products taking two at a time. But this very very exhaustive I want to know some another method as it will be proficient in higher powers. 
 [/INST] For any formal Laurent series $f(z) = \sum_{k=-\infty}^\infty a_k z^k$ in any variable $z$, let $[z^k]f(z)$ be a short hand for the coefficient $a_k$. 
Let $y = x^{-1}$, we can simplify the coefficient at hand as
$$\begin{align}\mathcal{C} \stackrel{def}{=} [x^{70}]\prod_{k=1}^{12}(x^k-k)
&= [x^{70}]x^{78} \prod_{k=1}^{12}(1-kx^{-k}) \\
&= [y^8]\prod_{k=1}^{12}(1-ky^k)
= [y^8]\prod_{k=1}^8(1-ky^k)\end{align}$$
We can split last product into two pieces
$$\begin{align}\prod_{k=1}^3 (1-ky^k) &= (1-y)(1-2y^2)(1-3y^3)
= (1-y-2y^2+2y^3)(1-3y^3)\\
&= 1-y-2y^2-y^3 + 3y^4 +6y^5 - 6y^6\\
\prod_{k=4}^8 (1-ky^k) &=
1 - 4y^4 - 5y^5 - 6y^6 - 7y^7 - 8y^8 + O(y^9)
\end{align}$$
Throwing away terms which doesn't contribute to $[y^8]$, we obtain
$$\begin{align}\mathcal{C} = &\; [y^8]\big((1-y-2y^2-y^3+3y^4)(1-4y^4-5y^5-6y^6-7y^7-8y^8)\big)\\
= &\; (1)(-8) + (-1)(-7) + (-2)(-6)+(-1)(-5)+(3)(-4)\\
= &\; -8 + 7 + 12 + 5 - 12\\
= &\; 4\end{align}$$
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