Closed formula of $(1+x_1)(1+x_1+x_2)^2\dots(1+x_1+\dots+x_n)^n$ I am working on a larger problem which I have managed to reduce to a pretty neat generating function. The answer will be given by the coefficient of $x_1x_2\dots x_n$ in 
$F(x_1,\dots,x_n)=(1+x_1)(1+x_1+x_2)^2\dots (1+x_1+x_2+\dots+x_n)^n$
Now, I have two questions about this function:


*

*I have been trying desperately to find a closed formula for this function, but without success. Am I missing something very obvious or is this as good as it gets?

*Is there a nice expression for the coefficient (call it $a_n$) of  $x_1x_2,\dots,x_n$? Here I might have made some progress. By writing each $(1+x_1+\dots+x_i)^i$ as a multinomial I have been able to come up with this expression (hopefully correct...):
\begin{align} 
\sum_{k_1+\dots+k_n=n}\left[\binom{1}{k_1}\binom{2}{k_2}\dots \binom{n}{k_n}(2-k_1)(3-(k_1+k_2))\dots(n-(k_1+\dots+k_{n-1})) \right]
\end{align}
but I don't seem to be able to reduce it to something nicer.


Any hints would be very much appreciated!
 A: To obtain the monomial $x_1x_2\cdots x_n$ from the product
$$
p = (1+x_1)(1+x_1+x_2)^2\cdots (1+x_1+\cdots+x_n)^n
$$
you have to pick the $x_n$ from one of the $(1+x_1+\cdots+x_n)$-factors and there are $n$ factors to choose from. Now you are left with the coefficient of $x_1x_2\cdots x_{n-1}$ in
$$
(1+x_1)(1+x_1+x_2)^2\cdots (1+x_1+\cdots+x_n)^{n-1}
$$
which is the same as the coefficient in
$$
(1+x_1)(1+x_1+x_2)^2\cdots (1+x_1+\cdots+x_{n-2})^{n-2}(1+x_1+\cdots+x_{n-1})^{2n-2}.
$$
Now you have to pick the $x_{n-1}$ from one of the $(1+x_1+\cdots+x_{n-1})$-factors and there are $2n-2$ to pick from.
Continuing this line of thought you get that the coefficient of $x_1x_2\cdots x_n$ in $p$ is
$$
n(2n-2)(3n-5)(4n-9)...(nn-*).
$$
Note that by construction the constants $0,2,5,9,\dots$ in this product are the sums
$$
0, 2, 2+3, 2+3+4, \dots
$$
which are given by
$$
\sum_{i=2}^k i = \frac{k(k+1)}{2} - 1 = \frac{(k-1)(k+2)}{2}.
$$
Hence, the desired coefficient is
$$
\prod_{k=1}^n \left(kn - \frac{(k-1)(k+2)}{2}\right).
$$
A: Write out all $n(n+1)/2$ factors in a row.
The $x_n$ comes from one of the final $n$ factors, the $x_{n-1}$ from one of the final $n+(n-1)$ factors but not the one used for $x_n$, and so on.
The product is
$$n(2n-2)(3n-5)(4n-9)(5n-14)...$$
But I can't help you with that. 
