Sum of partial factorials I was wondering if the following sum has a closed form:
$$S_n=\sum_{k=1}^{n-1} \frac{n!}{k!}=n+n\cdot(n-1)+\cdots+n!$$
$S_n$ satisfies the following recursive relation
$$S_n=n\cdot(S_{n-1}+1)$$
Is there a simple closed form representation of these sums?
 A: Edit: Initially I misread the summation, including an extra term.
This sequence is almost A007526 in OEIS, which has exponential generating function $$\frac{xe^x}{1-x}$$ and (more or less) closed form $a_n=\lfloor en!-1\rfloor$.
However, the sequence in OEIS is actually $$a_n=\sum_{k=0}^{n-1}\frac{n!}{k!}=S_n+n!\;,$$ which satisfies the same recurrence with $a_0=0$. For $S_n$ the recurrence starts at $S_1=0$ and holds thereafter. The $S_n$ are actually A038156 and satisfy $$S_n=\lfloor(e-1)n!)\rfloor-1\;.$$
A: Well it depends what you mean by a closed form..
$$S_n= n! [\frac{1}{1!}+..+\frac{1}{(n-1)!}] \,.$$
Then 
$$en!-S_n= n! [1-\frac{1}{n!}-\frac{1}{(n+1)!}-...]$$ 
$$en!-n!+2-S_n=1- n! [-\frac{1}{(n+1)!}-...]$$ 
Now, I think an easy computation shows that the RHS is between 0 and 1.
Thus
$$\lfloor en!-n!+2-S_n \rfloor =0$$
Hence
$$\lfloor en!-n!+2\rfloor =S_n$$
Is this a closed form or not? :)
P.S. Got to go, and typed in a hurry thus there might be mistakes in the computation. The basic idea should be right though...
