The value of $\quad \lim _{x \rightarrow 0} \dfrac{(1-x)^n-1}{x}=$

Let $f(x) = (1-x)^n$. The given limit is of the form $\lim_{x \to 0} \dfrac{f(x) - f(0)}{x - 0}$, which is the definition of the derivative $f'(0)$.

Calculate the derivative of $f(x) = (1-x)^n$ with respect to $x$:

$f'(x) = n(1-x)^{n-1} \times \dfrac{d}{dx}(1-x) = n(1-x)^{n-1} \times (-1) = -n(1-x)^{n-1}$

Evaluate the derivative at $x = 0$:

$f'(0) = -n(1-0)^{n-1} = -n(1)^{n-1} = -n$.

Answer: $-n$