TITLE: Closed formula for the sums $\sum\limits_{1 \le i_1 < i_2 < \dots < i_k \le n} i_1 i_2 \cdots i_k $?
QUESTION [7 upvotes]: I've worked out a few summation formulas, and I am hoping to find a pattern. Unless I have made a mistake somewhere, we have the following identities:
$$\sum_{1 \le i \le n} i = \frac{(n+1)n}{2} $$
$$\sum_{1 \le i < j \le n} ij = \frac{(3n+2)(n+1) \, n \, ( n-1)}{24}$$ 
$$\sum_{1 \le i < j < k \le n} ijk = \frac{(n+1)^2 \, n^2 \, (n-1)(n-2) }{48}$$ 
It seems clear that
$$p_k(n)= \sum_{1 \le i_1 < i_2 < \dots < i_k \le n} i_1 i_2 \cdots i_k $$ is a polynomial in $n$ of degree $2k$. But is there a nice closed-form formula for it? Maybe in terms of its factorization?
Comment: I realize that what I am looking for is the coefficient of $t^k$ in the expansion
$$(1+t)(1+2t) \dots (1+nt),$$
so maybe generating function techniques could be helpful. But the main way I know to extract the coefficient of $t^k$ is to take derivatives $k-1$ times and on the surface of things that looks like a huge mess.

REPLY [8 votes]: Starting from your generating function we have
$$p_k(n) = [t^k] \prod_{q=1}^n (1+qt).$$
This is $$[t^k] t^n \prod_{q=1}^n (1/t+q)
= [t^{k-n}]  \prod_{q=1}^n (1/t+q).$$
Set $t=1/v$ to get
$$[v^{n-k}]  \prod_{q=1}^n (v+q)
= [v^{n-k+1}]  \prod_{q=0}^n (v+q).$$
Now  the RHS  is precisely  the  generating function  of the  Stirling
numbers of the first kind and we get
$$\left[n+1\atop n+1-k\right].$$