If $f(x)=x^{-n}$ for $n\in \mathbb{N}$

Question

If $f(x)=x^{-n}$ for $n\in \mathbb{N}$ then prove

$f^{(k)}(x)=(-1)^{k} \frac{(n+k-1)!}{(k-1)!}$ $x^{-n-k} $

Of course I'm taking it by induction, but the main issue comes when $k+1$ and my final result comes to this

I derive $f^{(k)}(x)=(-1)^{k} \frac{(n+k-1)!}{(k-1)!}$ $x^{-n-k} $ therefore

$f^{(k+1)}(x)=(-1)^{k} \frac{(n+k-1)!}{(k-1)!}$ $x^{-n-k-1}(-n-k) =$

$f^{(k+1)}(x)=(-1)^{k+1} \frac{(n+k)!}{(k-1)!}$ $x^{-n-k-1}$

but my intuition tells me that I should get to

$f^{(k+1)}(x)=(-1)^{k+1} \frac{(n+k)!}{k!}$ $x^{-n-k-1}$

Any advice?

Answer 1

If I'm not mistaken, the derivative has a typo in the denominator. It should be $(n-1)!$ instead of $(k-1)!$: $$ f^{(k)}(x)=(−1)^k \frac{(n+k−1)!}{(n−1)!} x^{−(n−k)} . $$

Using this expression, the proof by induction should work.