在“已知m,n属于正自然数且1

1) (1+m)^n >(1+n)^m
2) (1+m)^n /(1+m)^(n-1)=1+m
3) (1+n)^m/(1+n-1)^m=(1+1/n)^m =1+C(1,m)/n +C(2,m)/n^2 +.+C(m,m)/n^m
m时恒立
(1+n)^m =[(1+n)/n]^m*[n/(n-1)]^m*[(n-1)/(n-2)]^m.[(m+2)/(m+1)]^m *(1+m)^m
《(1+m)*(1+m).(1+m) *(1+m)^m
==(1+m)^n
证毕