Răspuns :
Logica ar fi urmatoarea:
n!=1*2*3*...*(n-1)*n=(n-1)!*n.
[tex]Din~n!=(n-1)!*n=\ \textgreater \ (n-1)!= \frac{n!}{n} . \\ \\ Pentru~n=1,~avem:~(1-1)!= \frac{1!}{1}\ \textless \ =\ \textgreater \ 0!=1. [/tex]
n!=1*2*3*...*(n-1)*n=(n-1)!*n.
[tex]Din~n!=(n-1)!*n=\ \textgreater \ (n-1)!= \frac{n!}{n} . \\ \\ Pentru~n=1,~avem:~(1-1)!= \frac{1!}{1}\ \textless \ =\ \textgreater \ 0!=1. [/tex]