证：
∵A^2=A
∴对于任意正整数k,A^k=A
根据二项式展开【C(n,k)代表组合数】
(A+I)^m
=C(m,0)[A^m]+C(m,1)[A^(m-1)]+C(m,2)[A^(m-2)]+……+C(m,m)[I]
=C(m,0)[A]+C(m,1)[A]+C(m,2)[A]+……+C(m,m-1)[A]+C(m,m)[I]
=[C(m,0)+C(m,1)+……+C(m,m-1)][A]+I
∵C(m,0)+C(m,1)+……+C(m,m)=(1+1)^m=2^m
∴C(m,0)+C(m,1)+……+C(m,m-1)=2^m-C(m,m)=2^m-1
∴(A+I)^m
=[C(m,0)+C(m,1)+……+C(m,m-1)][A]+I
=(2^m-1)A+I
证毕!