Is $n! \sum_{i=0}^n{\frac{(-1)^i}{i!}}- (n-1)! \bigg[\sum_{i=0}^{n-2}{\frac{(-1)^i}{i!}}+...+\sum_{i=0}^{2}{\frac{(-1)^i}{i!}}\bigg]=(n-1)!$ true? I am in the middle of doing a problem and has this sort of expression. I have a feeling that the following equality holds:
$$n! \sum_{i=0}^n{\frac{(-1)^i}{i!}}- (n-1)! \bigg[\sum_{i=0}^{n-2}{\frac{(-1)^i}{i!}}+\sum_{i=0}^{n-3}{\frac{(-1)^i}{i!}}+\sum_{i=0}^{n-4}{\frac{(-1)^i}{i!}}+...+\sum_{i=0}^{2}{\frac{(-1)^i}{i!}}\bigg]=(n-1)!$$
Does the above identity hold? 
If it is, how we can show it?
Any help would be highly appreciated!
 A: You can rewrite your LHS this way: $$LHS = n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{j=2}^{n-2} \sum_{i=0}^j \frac{(-1)^i}{i!}\right) \\
= n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{j=0}^{n-2} \sum_{i=0}^j \frac{(-1)^i}{i!} - 1\right)$$
Swapping sums gives:
$$LHS = n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{i=0}^{n-2} \sum_{j=i}^{n-2} \frac{(-1)^i}{i!} - 1\right) \\ 
= n! \sum_{i=0}^n \frac{(-1)^i}{i!} - (n-1)!\left(\sum_{i=0}^{n-2} (n-1-i) \frac{(-1)^i}{i!} - 1\right) \\
= n! \sum_{i=0}^n \frac{(-1)^i}{i!} - \sum_{i=0}^{n-2} (n-1)!(n-1-i) \frac{(-1)^i}{i!} + (n-1)! \\
= (-1)^n (1 - n) + n! \sum_{i=0}^{n-2} \frac{(-1)^i}{i!} - \sum_{i=0}^{n-2} (n-1)!(n-1-i) \frac{(-1)^i}{i!} + (n-1)! \\
= (-1)^n (1 - n) + \sum_{i=0}^{n-2} \frac{(-1)^i}{i!} \left[ n! - (n-1)!(n-1-i) \right] + (n-1)! \\
= \underbrace{(-1)^n (1 - n) + \sum_{i=0}^{n-2} \frac{(-1)^i}{i!}(n-1)!(1+i)}_A + (n-1)!$$
Can you now show that $A = 0$ ?
A: Note that this identity would be equivalent to the following. If the original expression is true then the following holds:
$ \frac{\Gamma(n+1,-1)}{\Gamma(n)} = e+\sum_{k=2}^n {\frac{\Gamma(k+1,-1)}{\Gamma(k+1)}}$
I am not sure, however, if the original identity in the question is true.
A: Indeed the identity is valid. Dividing OPs expression by $n!$ and putting the bracketed sum to the right we show

The following is valid for $n> 0$
  \begin{align*}
\sum_{i=0}^n\frac{(-1)^n}{i!}=\frac{1}{n}\left(1+\sum_{j=2}^{n-2}\sum_{i=0}^j\frac{(-1)^i}{i!}\right)
\end{align*}

$$ $$

We obtain starting with the RHS
  \begin{align*}
\frac{1}{n}&\left(1+\sum_{j=2}^{n-2}\sum_{i=0}^j\frac{(-1)^i}{i!}\right)\\
&=\frac{1}{n}\sum_{j=0}^{n-2}\sum_{i=0}^j\frac{(-1)^i}{i!}\tag{1}\\
&=\frac{1}{n}\sum_{i=0}^{n-2}\sum_{j=i}^{n-2}\frac{(-1)^i}{i!}\tag{2}\\
&=\frac{1}{n}\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}(n-i-1)\tag{3}\\
&=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}-\frac{1}{n}\sum_{i=1}^{n-2}\frac{(-1)^i}{(i-1)!}-\frac{1}{n}\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}\tag{4}\\
&=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}+\frac{1}{n}\sum_{i=0}^{n-3}\frac{(-1)^i}{i!}-\frac{1}{n}\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}\tag{5}\\
&=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}-\frac{1}{n}\frac{(-1)^{n-2}}{(n-2)!}\tag{6}\\
&=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}+\frac{(-1)^{n-1}}{n}(n-1)\tag{7}\\
&=\sum_{i=0}^{n-2}\frac{(-1)^i}{i!}+\frac{(-1)^{n-1}}{(n-1)!}+\frac{(-1)^n}{n!}\\
&=\sum_{i=0}^{n}\frac{(-1)^i}{i!}\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}
  and the claim is finished.

Comment:


*

*In (1) we observe that $$1=\sum_{j=0}^{1}\sum_{i=0}^j\frac{(-1)^i}{i!}$$

*In (2) we exchange the sums by noting that $$0\leq i\leq j\leq n-2$$

*In (3) we see the inner sum indexed with $j$ is $n-i-1$

*In (4) we split the sum according to $n-1-i$ and observe that in the middle sum the term  with $i=0$ vanishes

*In (5) we shift the index of the middle sum by one and prepare so for telescoping

*In (6) we use telescoping

*In (7) we expand with $\frac{n-1}{n-1}$
