How do you simplify the factorial expression #((n+1)!)/(n!)#?

1 Answer
Alan N.
Feb 18, 2017

#n+1#

Explanation:

#(n+1)! = (n+1) * n * (n-1 )* (n-2)* ... 2*1 #

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

#:. ((n=1)!)/(n!) = ((n+1) * cancel( n * (n-1 )* (n-2)* ... 2*1))/cancel( n * (n-1 )* (n-2)* ... 2*1)#

#=n+1#

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