Prove that $ \frac{n(n-1)(n-k+1)}{k(k-1)\cdots1} = \frac{n!}{(n-1)!k!}$

Question

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

Proof:

This is what I have tried:

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

$\frac{n!}{k!} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k(k-1)\cdots1}$

This is where my attempt ends.

Are my steps correct? If not how can I improve them?

New Questions:

How is it that n-k magically appears on the right side? I get that n-k and 3 times times 2 times 1 cancels out in the numerator and the denominator which leaves

$\frac{n(n-1)(n-2)...(n-k+1)}{k(k-1)....1}$ which is equivalent to ${n \choose k}$ or \frac{n!}{(n-k)!k!}....is that correct?

Answer 1

This is what I think is required to be proved

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

If yes, then,

$\begin{array}{c} \frac{{n(n - 1)(n - k + 1)}}{{k(k - 1) \cdots 1}} = \frac{{n(n - 1)(n - k + 1)\left( {n - k} \right)...3.2.1}}{{\left( {k(k - 1) \cdots 1} \right)\left( {\left( {n - k} \right)...3.2.1} \right)}}\\ = \frac{{n!}}{{\left( {k!} \right)\left( {n - k} \right)!}} \end{array}$

Answer 2

It is not correct as stated: if we let $n=k>2$ then we have on the LHS: $\frac{k(k-1)(1)}{k!}=\frac{1}{(k-2)!}$ and for the RHS: $\frac{1}{(k-1)!}$.