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!

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.
