What would be the solution to the $k^{th}$ derivative of the following function
$$\dfrac{1}{x^y}$$ With respect to $x$ where y is a constant.
I have calculated the first derivative $$-y\dfrac{1}{x^{y+1}}$$ and the second $$y(y+1)\dfrac{1}{x^{y+2}}$$ third $$-y(y+1)(y+2)\dfrac{1}{x^{y+3}}$$ fourth $$y(y+1)(y+2)(y+3)\dfrac{1}{x^{y+4}}$$
Therefore the nth derivative is $$(-1)^n (?) \dfrac{1}{x^{y+n}} $$
I cannot seem to spot the patten and cannot get the $k^{th}$ derivative.
$$ \begin{align} \displaystyle \frac{\partial^k}{\partial x^k} \frac 1{x^y} &= (-1)^k \displaystyle \frac {(y + (k - 1))!}{(y-1)!} \displaystyle \frac 1{x^{y+k}} \\ &= (-1)^k \binom{y + (k-1)}{k} k! \displaystyle \frac 1{x^{y+k}} , \text{where } k = 1, 2, 3, ... \end{align}$$