Question 897931
By definition


n! = n*(n-1)*(n-2)*(n-3)...(3)*(2)*(1)


We can rewrite "(n-1)*(n-2)*(n-3)...(3)*(2)*(1)" as simply "(n-1)!"


So, n! = n*(n-1)!


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n!/(n-1)!=5


[n*(n-1)!]/(n-1)!=5


n = 5