I have a question based on Irreducible and Connected Permutations.
I was able to use the notion of connected permutations to construct a combinatoric proof for
\begin{equation} \sum_{i=1}^{n}c_{i}(n-i)!=n!. \end{equation}
Now let $\displaystyle F(x)=\sum_{n \geq 1} n! x^{n}$ and $\displaystyle G(x)=\sum_{n \geq 1} c_{n} x^{n}$.
I was then asked to prove that $1-G(x)$ and $1+F(x)$ are multiplicative inverses. It would seem to be an easy question but I cannot seem to see my mistake:
\begin{equation} \begin{aligned} (1-G(x))(1+F(x)) &= 1 \\ 1 - G(x) + F(x) - G(x)F(x) &= 1 \\ 1 - \sum_{n\geq 1}c_n x^n + \sum_{n \geq 1} n! x^{n} - \sum_{n \geq 1} \left[ \sum_{i=1}^{n}c_{i}(n-i)! \right] x^{n} &= 1 \\ 1 - \sum_{n\geq 1}c_n x^n + \sum_{n \geq 1} n! x^{n} - \sum_{n \geq 1}n! x^{n} &= 1 \\ 1 - \sum_{n\geq 1}c_n x^n &= 1 \\ \sum_{n\geq 1}c_n x^n &= 0 \\ \end{aligned} \end{equation}
This would mean that every coefficient of $G(x)$ is zero, which is absurd.
Note that I substituted the coefficient of $G(x)F(x)$ with the combinatoric identity. I have approached the problem in many different ways (e.g. changing indices, constructing the inverse of $F(x)$ and comparing to $G(x)$) and I ultimately come to the same result.
I would appreciate if anyone could point out my error(s).
Thank you very much!
Let $a_0=1$ and $a_n=-c_n$ for $n\ge 1$; then
$$\begin{align*} \big(1-G(x)\big)\big(1+F(x)\big)&=\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}n!x^n\right)\\\\ &=\sum_{n\ge 0}\sum_{k=0}^na_k(n-k)!x^n\\\\ &=1\;, \end{align*}$$
since
$$\sum_{k=0}^na_k(n-k)!=\begin{cases} 1,&\text{if }n=0\\\\ n!-\sum\limits_{k=1}^nc_k(n-k)!=n!-n!=0,&\text{if }n\ge 1\;. \end{cases}$$