Show that $V_n^k=\frac{n!}{(n-k)!}$

Question

Show that $$V_n^k=\dfrac{n!}{(n-k)!}.$$ The general formula for variations is $V_n^k=n(n-1)(n-2)...(n-k+1).$ We have $$\dfrac{n!}{(n-k)!}=\dfrac{n!}{(n-k)(n-k-1)...2.1}.$$ What can I do next?

Answer 1

Write the numerator as follows and cancel out terms: $$\frac{n(n-1)\dots(n-k+1)(n-k)\dots(2)(1)}{(n-k)(n-k-1) \dots (2)(1)} = n(n-1)(n-2)\dots(n-k+1)=V_n^k$$