Searching for a "good formula" for "$\sum_{i=0}^{n-1}\prod_{j=0}^{i} (n-j)$ I m trying to find a "good formula" for
$$\sum_{i=0}^{n-1}\prod_{j=0}^{i}(n-j)=n+n(n-1)+\dots+n(n-1)\dots 3\cdot 2+n!$$
In fact, this question has been asked before: Expression for $n+n(n-1)+n(n-1)(n-2)+...+n!$  but I'm searching for a "good formula", namely a closed expression for this sum depending from a fixed number of elementary operations (sums and differences, products and divisions, factorials $\dots$ ) involving algebraic quantities.
I'm asking for such a solution of the problem because, even if $e$ is actually one of the most important mathematical constants, the idea behind the answer of the previous topic is "Ok, I prove that a particular series related to the problem is convergent and I express all thanks to this result", which is obviously correct and probably the best way to solve such problems, but it isn't a solution in the sense I've previously explained (although very meaningful).
Now I'll share with you some interesting results and points of view : let's define $(a_n)_{n\in \mathbb{N}}$ the sequence such that
$$ a_0=0 \,\,\text{and for}\,\, n\geq 1 \,\,\text{we have}\,\,a_{n}=\sum_{i=0}^{n-1}\prod_{j=0}^i (n-j)$$
You can obviously write $a_{n+1}=(n+1)(1+a_n)$ and try to bash this recurrence relation using the theory of Generating Functions, but, as expected, you get quite easily involved in exponentials and quite hard integrals, which if solved will probably give you the expression already known for $a_n$.
So, perhaps, a new point of view has to be found. For example, I've thought to consider the sequence $(p_n)_{n\in\mathbb{N}}$ of polynomials such that
$$\begin{cases}
p_0 \,\,\,\text{is the identically null polynomial}\\
p_n(x)=\sum_{i=0}^{n-1}\prod_{j=0}^i (x-j)
\end{cases}$$
All these polynomial are such that $p_n(n)=a_n$ and $p_n$ has degree $n$ when $n\geq 1$. You can easily see that for all $n\geq 1$ we have $$\forall k<n: p_n(k)=a_k$$
and so we have to find a new particular value for $p_n$ in order to fix it with the previous values of $a_k$. So we use the fact that
$$p_n(-1)=\sum_{j=1}^{n}((-1)^j\cdot j!)$$
Since a polynomial of degree $n$ with coefficients in a field is fixed by $n+1$ values, we have that
$$p_n(x)=\frac{x(x-1)\dots (x-n+1)}{(-1)^n \cdot n!}\cdot \sum_{j=1}^{n}((-1)^j j!)+\sum_{k=0}^{n-1}\frac{a_k\cdot \prod_{j=-1\\j\neq k}^{n}(x-j)}{(k+1)!\cdot (-1)^{n-1-k}(n-1-k)!}$$
And so
$$a_n=p_n(n)=(-1)^n\sum_{j=1}^{n}((-1)^j j!)+\sum_{k=0}^{n-1}\Big[(-1)^{n-1-k}\binom{n+1}{k+1}a_k\Big]$$
namely
$$\sum_{k=1}^{n}\Big[(-1)^{n+1-k}\cdot k!\Big]=\sum_{k=1}^{n}\Big[(-1)^{n+1-k}\binom{n+1}{k+1}a_k\Big]$$
which has got little simmetry but perhaps it's only my imagination and, perhaps, all these ravings are completely useless. However, thank you for the attention.
 A: I don't think you'll find any simplified version of this involving some low-degree polynomial. At best, your expression is equal to
$en\Gamma(n,1)=en\int_1^\infty t^{n-1}e^{-t}dt$
where $\Gamma(s,x)$ is the incomplete gamma function.
A: I don't think there's any reason to believe such a formula exists. What we can say is that this expression is equal to
$$f(n) = n! \left( \sum_{i=0}^{n-1} \frac{1}{i!} \right)$$
and the expression in parentheses converges very rapidly to $e$. In fact we have
$$e - \frac{f(n)}{n!} = \sum_{i=n}^{\infty} \frac{1}{i!} \le \frac{1}{n!} \sum_{i=0}^{\infty} \frac{1}{(n+1)^i} = \frac{1}{n!} \left(\frac{n+1}{n} \right)$$
so
$$1 \le e \cdot n! - f(n) < 1 + \frac{1}{n}.$$
This is very small error, and for $n \ge 1$ it implies that we must in fact have
$$\boxed{ f(n) = \lfloor e \cdot n! \rfloor - 1}.$$
I think this is as good as it gets. Edit: I see that this was already given in the linked answer. I guess I'll just second the linked answer: I don't think it gets better than this.
A: The product is the definition of the Falling Factorial.
$$
\prod\limits_{j = 0}^{m - 1} {\left( {n - j} \right)}
  = n^{\,\underline {\,m\,} }  = m!\left( \matrix{  n \cr  m \cr}  \right)
 = {{n!} \over {\left( {n - m} \right)!}}
$$
Unfortunately there is not a closed form for the sum of them over the "exponent", similar to the geometric sum.
The relation between the Falling Factorials and the powers of $n$ is expressed through the
Stirling N. of 1st kind as
$$
n^{\,\underline {\,m\,} }  = \sum\limits_{0\, \le \,k\, \le \,m}
   {\left( { - 1} \right)^{\,m - k} \left[ \matrix{ m \cr   k \cr}  \right]n^{\,k} } 
$$
So your sum is
$$
S(n) = \sum\limits_{0\, \le \,m\, \le \,n} {n^{\,\underline {\,m\,} } }
  = \sum\limits_{0\, \le \,k\, \le \,n} {\left( {\sum\limits_{k\, \le \,m\, \le \,n}
 {\left( { - 1} \right)^{\,m} \left[ \matrix{  m \cr   k \cr}  \right]} } \right)
\left( { - 1} \right)^{\,k} n^{\,k} } 
$$
$S(n)$ is the OEIS seq. A000522 where you can find many other properties and relations.
A: Using Pochhammaer symbols $$\prod_{j=0}^{i} (n-j)=(-1)^i n (1-n)_i$$ Using the incomplete gamma function
$$\sum_{i=0}^{n-1}\prod_{j=0}^{i} (n-j)=e\, n \,\Gamma (n,1)$$
