Factorial identity $n!=1+(1-1/1!)n+(1-1/1!+1/2!)n(n-1)+\cdots$ Show that $\displaystyle{n!=1+\left(1-\frac1{1!}\right)n+\left(1-\frac1{1!}+\frac1{2!}\right)n(n-1)+\cdots}$. I can't figure out how this can be solved. I tried to use the binomial theorem but I couldn't prove it. Any help will be greatly appreciated.
 A: If you divide by $n!$ then the requested identity is
$$1=\sum_{k=0}^n\frac1{k!}\sum_{j=0}^{n-k}\frac{(-1)^j}{j!},$$
perhaps better written using the Iverson bracket as
$$1=\sum_{j,k}\frac{(-1)^j}{j!k!}\bigl[j\ge0,k\ge0,j+k\le n\bigr].$$
(The bracket takes the value $1$ if the statement inside holds, and $0$ otherwise.)
Notice that $$\sum_{j,k}\frac{(-1)^j}{j!k!}\bigl[j\ge0,k\ge0,j+k=m\bigr]
=\begin{cases}\dfrac{(1-1)^m}{m!}=0&m\ge1,\\1&m=0\end{cases}$$ 
Now sum this over $m\in\{0,1,\ldots,n\}$ to get the desired result.
A: Assuming the sum is supposed to terminate after $n$ terms, your sum is:
$$
\frac{n!}{n!} \sum_{k=1}^{n+1} \sum_{m=1}^k \frac{(-1)^{m-1}}{(m-1)!}\prod_{r=0}^{k-2} (n-r)
$$
$$
=n!\sum_{k=1}^{n+1} \sum_{m=1}^k \frac{(-1)^{m-1}}{(m-1)!} \frac{\prod_{r=0}^{k-2} (n-r)}{n!}
$$
$$
=n!\sum_{k=1}^{n+1} \sum_{m=1}^k \frac{(-1)^{m-1}}{(m-1)!(n+1-k)!}
$$
$$
=n!\sum_{k=0}^{n} \sum_{m=0}^{k} \frac{(-1)^{m}}{m!(n-k)!}
$$
$$
=n!\sum_{k=0}^{n} \frac{1}{(n-k)!} \sum_{m=0}^{k} \frac{(-1)^{m}}{m!}
$$
$$
=\frac{1}{e}\sum_{k=0}^{n} \binom{n}{k}{\Gamma(k+1,-1)}
$$
But I'm already just quoting the incomplete Gamma function which is essentially just rewriting the sum, I can't show this is $n!$.
A: Here is a combinatorial argument. Rewrite the sum like this
$$0!\times 1 \color{#F00}{+} 1! (1-1/1!) n/1! \color{#F00}{+} 2! (1-1/1!+1/2!) n(n-1)/2! 
\\\color{#F00}{+} 3! (1-1/1!+1/2!-1/3!) n(n-1)(n-2)/3!+\ldots$$
Two observations: first the term $$q! \times (1-1/1!+1/2!-1/3!+\cdots\pm 1/q!)$$
counts derangements on permutations of $q$ elements and second the term $n(n-1)\cdots(n-q)/q!$ is ${n\choose q} = {n\choose n-q}$ says to choose $n-q$ elements from $n$ elements.

Therefore the sum is a classification of the set of permutations by the number of fixed points: choose $n-q$ fixed points and combine with a derangement on $q$ elements. There are $n!$ permutations however. Done.
