what is the n-k derivative of $x^n$? Also, why is $n!/k! = ...$

Question

I am having troubles finding $\frac{d^{n-k}x^n}{dx^{n-k}}$ where $ k \leq n$ I believe it is equal to $n(n-1)(n-2)....k(k+1)x^k$ but htis is just from obersation, I do not know why it's that exactly.

Also, I am having troubles justifying htat $\frac{n!}{k!} = n(n-1)(n-2)....k(k+1)$

Answer 1

Start from $$(x^n)^{(k)}=\dfrac{n!}{(n-k)!}x^{n-k},\quad\text{which is easy to prove by induction.}$$ and apply it replacing $k$ with $n-k$: $$(x^n)^{(n-k)}=\dfrac{n!}{(n-(n-k))!}\,x^{n-(n-k)}=\dfrac{n!}{k!}\,x^{k}. $$

Also $$\frac{n!}{k!}=\frac{n(n-1)\cdots(k+1)k(k-1)\cdots1}{k(k-1)\cdots1}=n(n-1)\cdots(k+1).$$

Answer 2

$\require{cancel}$

$$\frac{n!}{k!} = \frac{n(n-1)(n-2)\dots}{k(k-1)(k-2)\dots} = \frac{n(n-1)(n-2)\dots(k+1)k(k-1)(k-2)\dots}{k(k-1)(k-2)\dots} = \frac{n(n-1)(n-2)\dots(k+1)\cancel{k(k-1)(k-2)\dots}}{\cancel{k(k-1)(k-2)\dots}} = n(n-1)(n-2)\dots(k+1)$$