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 $?

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

$$\left[n+1\atop n+1-k\right].$$

$$p_{k}(n) = S_1(n+1, n-k+1)$$ where $S_1$ are the unsigned Stirling numbers of the first kind.

EDIT: We then have $$p_k(x) = {x \choose k} S_k(x)$$ where $S_k(x)$ are the Stirling polynomials and $${x \choose k} = \dfrac{1}{k!} \prod_{j=0}^{k-1} (x - j)$$