(1)由a1=2,得a2=a12-a1+1=3由a2=3,得a3=a22-2a2+1=4由a3=4,得a4=a32-3a3+1=5(1)用数学归纳法证明①由a1=2=1+1知n=1时,an=n+1成立设n=k(k属于正整数)时an=n+1成立,即ak=k+1则当n=k+1时,因为an+1=an2-nan+1,所以ak+1=ak2-k(k+1)+1=(k+1)2-k(k+1)+1=k2+2k+1-k2-k+1=k+2综上,an=n+1成立