Answer
$n(n-1)...(n-k+1)\ or\ \prod_{i=n-k+1}^{n}(i)$
Work Step by Step
$\frac{n!}{(n-k)!}=\frac{n(n-1)...(n-k+1)(n-k)!}{(n-k)!}=n(n-1)...(n-k+1)=\prod_{i=n-k+1}^{n}(i)$
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