Can this sum be simplified: $ \sum_{k=0}^{n-1} { n -1 \choose k } (-2)^{k} (2n - k)! $? Can this expression be further simplified : $ \sum_{k=0}^{n-1} { n -1 \choose k } (-2)^{k} (2n - k)! $? This is the coefficient of $x^{2n}$ in the formal power series expansion of $(1-2x)^{n-1} \times \sum_{ k \geq 0} k! x^k$. 
Motivation: I came across this when trying to solve a problem using inclusion-exclusion principle, I am not mentioning the original problem because I am interested in this sum as an independent problem.

As a way of restating Bill's answer, what you have can in fact be expressed in terms of the so-called (generalized) Bessel polynomial:
$$y_n(x;a)=(n+a-1)_n \left(\frac{x}{2}\right)^n {}_1 F_1 \left(-n;-2n-a+2;\frac{2}{x}\right)$$
(the Kummer hypergeometric series degenerates to a polynomial here because both the numerator and denominator parameters are negative integers)
Your original expression, then, in terms of the (generalized) Bessel polynomial, is
$$(-2)^{n-1}(n+1)!y_{n-1}(-1;4)$$
The references in the DLMF can point you to papers where the (generalized) Bessel polynomials have been studied; for an elementary treatment, see Chihara's An Introduction to Orthogonal Polynomials.
