<s>[INST] Sum of magnitudes of coefficients of polynomial $(x-1)(x-2)(x-3)\cdots(x-(n-1))$ The title says most of it. I have found that the sum of the coefficients of the polynomial$(x-1)(x-2)(x-3)\cdots(x-(n-1))$ yields $n!$. 
For example, the coefficients of the polynomial $(x-1)(x-2)$ sum to $1+3+2=6=3!$ and similarly the coefficients of $(x-1)(x-2)(x-3)$ sum to $1+6+11+6=24=4!$. 
My question, then is how might I prove this more generally. I have the feeling that induction might be an optimal way to go about this, but am unsure of the specifics. Any help is appreciated!
 [/INST] Since all roots of the polynomial are positive, Descartes' Rule of signs guarantees that the signs of the coefficients alternate, for example with $n=4$:
$(x-1)(x-2)(x-3)=x^3\color{purple}{-}6x^2\color{purple}{+}11x\color{purple}{-}6$
Because of this sign pattern the absolute values of the coefficients are recovered by putting $x=-1$ so that the powers of $x$ correlate with the alternating signs of the coefficients:
$(-1-1)(-1-2)(-1-3)=(-1)^3-6(-1)^2+11(-1)11x-6=-(1+6+11+6)$
and upon taking absolute values
$(1+1)(1+2)(1+3)=1+6+11+6$
where for the general case the product on the left is $n!$ and the sum on the right contains the desired coefficient magnitudes.
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