Answer
$ n(n-1)...(n-k+2)\ or\ \prod_{i=n-k+2}^{n}(i)$
Work Step by Step
$\frac{n!}{(n-k+1)!}=\frac{n(n-1)...(n-k+2)(n-k+1)!}{(n-k+1)!}=n(n-1)...(n-k+2)=\prod_{i=n-k+2}^{n}(i)$
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