To compute the derivative $\left. \frac{dy}{dx} \right|_{x=1}$ where $y = (1-x)(2-x)\dots(n-x)$, apply the product rule to $n$ functions $f_k(x) = k - x$, each with derivative $f_k'(x) = -1$.

The derivative is given by summing over terms where each factor is differentiated in turn:

$\frac{dy}{dx} = \sum_{k=1}^{n} \left(\prod_{j \ne k} (j - x)\right) \cdot (-1)$

Evaluating at $x = 1$, note that $(1 - x) = 0$ when $x = 1$. Every term except the first contains the factor $(1 - x)$ and therefore vanishes.

The surviving first term is:

$(-1) \cdot (2-1)(3-1)\dots(n-1) = (-1) \cdot 1 \cdot 2 \cdots (n-1) = -(n-1)!$

Hence, $\left. \frac{dy}{dx} \right|_{x=1} = -(n-1)!$, corresponding to choice C.