How can the factorial of 0 be 1?

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Answer 1 · 2015-06-10T15:50:14

If you know the value of #n!# then you can calculate #(n-1)!# as
#(n!)/n#; since #1! =1# then #0! = (1-1)! = 1/1 = 1#

Actually Nelson's answer if probably correct, but there is some justification for the definition.

Answer 2 · 2015-06-10T15:58:16

Because there is one permutation of zero objects.

Interpreting #n!# to be the number of permutations of #n# objects.

And agreeing that there is a permutation of #0# object (namely the empty permutation).

Leads one to state that #0! = 1#

Answer 3 · 2015-06-11T21:16:50

We can do some MATH and STUFF!

At #n_0 = 1#:
#(n!)/((n+1)!) = (1*cancel(2*3*4*...*n))/(cancel(2*3*4*5*...*n*)(n+1))#

#= 1/(n+1)#

If #(n!)/((n+1)!) = 1/(n+1) = (0!)/(1!)#, then, with #n = 0#:

#0! = 1!*(1/(0+1)) = (1!)/(1) = 1#