Answer
$n! = \underline{n(n-1)(n-2)(n-3)...(3)(2)(1)}$
$0! = \underline{1}$
Work Step by Step
RECALL:
(i) For any counting (or natural) number $n$,
$n! = n(n-1)(n-2)(n-3)...(3)(2)(1)$
(ii) $0! = 1$.
Therefore
$n! = n(n-1)(n-2)(n-3)...(3)(2)(1)$ and $0! = 1$.
