设$y=uv$，则
$$y'=u'v+uv'，$$
$$y''=\left( u'v+uv' \right)'=u''v+2u'v'+uv''，$$
$$y'''=\left( u''v+2u'v'+uv'' \right)'=u'''v+3u''v'+3u'v''+uv'''，$$
　　如此下去，不难看到，计算结果与二项式$\left( u+v \right)^n$展开式极为相似，用数学归纳法，可得
$$\left( uv \right)^{\left( n \right)}=u^{\left( n \right)}v^{\left( 0 \right)}+C_n^1u^{\left( n-1 \right)}v^{\left( 1 \right)}+C_n^2u^{\left( n-2 \right)}v^{\left( 2 \right)} + \cdots + C_n^ku^{\left( n-k \right)}v^{\left( k \right)} + \cdots + u^{\left( 0 \right)}v^{\left( n \right)}$$$$=\sum\limits_{k = 0}^nC_n^ku^{\left( n-k \right)}v^{\left( k \right)}，$$
　　其中$u^{\left( 0 \right)}=u$，$v^{\left( 0 \right)}=v$。