The sequence $\left\{ a _ { n } \right\}$ satisfies $a _ { 1 } = 1$, and with $S _ { n } = \sum _ { k = 1 } ^ { n } a _ { k }$, $$a _ { n + 1 } = ( n + 1 ) S _ { n } + n ! \quad ( n \geq 1 )$$ The following is the process of finding the general term $a _ { n }$.
For a natural number $n$, since $a _ { n + 1 } = S _ { n + 1 } - S _ { n }$, by the given equation, $$S _ { n + 1 } = ( n + 2 ) S _ { n } + n ! \quad ( n \geq 1 )$$ Dividing both sides by $( n + 2 ) !$, $$\frac { S _ { n + 1 } } { ( n + 2 ) ! } = \frac { S _ { n } } { ( n + 1 ) ! } + \frac { 1 } { ( n + 1 ) ( n + 2 ) }$$ Let $b _ { n } = \frac { S _ { n } } { ( n + 1 ) ! }$. Then $b _ { 1 } = \frac { 1 } { 2 }$ and $$b _ { n + 1 } = b _ { n } + \frac { 1 } { ( n + 1 ) ( n + 2 ) }$$ Finding the general term of the sequence $\left\{ b _ { n } \right\}$, $$b _ { n } = \frac { ( \text{(가)} ) } { n + 1 }$$ Therefore, $$S _ { n } = \text{(가)} \times n!$$ Thus, $$a _ { n } = \text{(나)} \times ( n - 1 ) ! \quad ( n \geq 1 )$$ When the expressions that fit (가) and (나) are $f ( n )$ and $g ( n )$ respectively, what is the value of $f ( 7 ) + g ( 6 )$? [4 points]
(1) 44
(2) 41
(3) 38
(4) 35
(5) 32