Equality of 2 forms of summation - derangement problem The number of ways of placing $n$ objects not in position is given by the inclusion-exclusion number $D_n$: 
$n! \left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}+....+(-1)^n\dfrac{1}{n!} \right)$
which can also be written as:
$n!\sum_{i=1}^{n-1} (-1)^{i+1}/(i+1)!$
Using a different approach, I came to the following answer, which has probably already been proven:
$(n-1)!\sum_{i=1}^{n-1} (-1)^{i+1}(n-i)/(i-1)!$
The 2 different summations give the same answer ( I have tested for n=2,3,4,5,6,7).  However,each term for any $i$ is different in each summation.  How can the equality of the 2 summations be proved (if it is indeed he case for all $n$)?  
 A: For $n=1$ both expressions are zero. For $n\ge 2$ we start with the second expression, shift the index by one, and increase the upper limit (which adds nothing to the sum):
$$
(n-1)!\sum_{i=1}^{n-1} (-1)^{i+1}\frac{n-i}{(i-1)!} 
= (n-1)!\sum_{i=0}^{n-2} (-1)^i\frac{n-i-1}{i!} 
= (n-1)!\sum_{i=0}^{n-1} (-1)^i\frac{n-i-1}{i!} 
$$
Now split the sum into three parts:
$$
= (n-1)!\left( n \sum_{i=0}^{n-1} \frac{(-1)^i}{i!} -  \sum_{i=0}^{n-1} \frac{(-1)^i i}{i!} - \sum_{i=0}^{n-1} \frac{(-1)^i}{i!} \right) 
$$
In the middle sum the term for $i=0$ is zero, and in all other terms a factor $i$ can be canceled:
$$
= (n-1)!\left( n \sum_{i=0}^{n-1} \frac{(-1)^i}{i!} -  \sum_{i=1}^{n-1} \frac{(-1)^i}{(i-1)!} - \sum_{i=0}^{n-1} \frac{(-1)^i}{i!} \right)  \\
= n! \sum_{i=0}^{n-1} \frac{(-1)^i}{i!} - (n-1)! \left(\sum_{i=1}^{n-1} \frac{(-1)^i}{(i-1)!}  +  \sum_{i=0}^{n-1} \frac{(-1)^i}{i!}\right) 
$$
Now shift the indices in the middle sum:
$$
= n! \sum_{i=0}^{n-1} \frac{(-1)^i}{i!} - (n-1)! \left(\sum_{i=0}^{n-2} \frac{(-1)^{i+1}}{i!}  +  \sum_{i=0}^{n-1} \frac{(-1)^i}{i!}\right)
$$
and note that in the second and third sum all terms cancel, with the only exception of the $i=n-1$ term in the third sum, so that the expression is finally equal to
$$ \\
=  n! \sum_{i=0}^{n-1} \frac{(-1)^i}{i!} + (-1)^n = 
n! \left( 1-\dfrac{1}{1!}+\dfrac{1}{2!}+....+(-1)^n\dfrac{1}{n!} \right)
$$
and that is the desired equality.
A: Let $LHS(n)$ denote the expression $n!\sum_{i=1}^{n-1} (-1)^{i+1}/(i+1)!$
Let $RHS(n)$ denote the expression $(n-1)!\sum_{i=1}^{n-1} (-1)^{i+1}(n-i)/(i-1)!$
Claim $LHS(n)=RHS(n) \ \ \forall n$
E.g. for $n=4$ we have that $LHS=12-4+1=9$ and $RHS=18-12+3=9$
Proof by Induction:
Assume that $LHS(n)=RHS(n)$ for some given $n$
Want $LHS(n+1)=RHS(n+1)$
$LHS(n+1)=\dots=(n+1)LHS(n)+(-1)^{n+1}$
$RHS(n+1)=\dots$
THAT IS ALL I HAVE TIME FOR RIGHT NOW, SORRY!
