Find coefficients of the polynomial $1-(1-x)(1-2x)\dotsb (1-nx)$ As in the title I'm finding the coefficients $a_k$ (coefficient of $k$th term) of the following polynomial
$$p(x)=1-\prod_{k=1}^{n}(1-kx)$$
As one clearly sees $a_0=0$ and, with a bit of computation that $a_1=\frac{n(n+1)}{2}$.
For $a_2$ the question is more complicated, I think the answer is $$a_2=\frac{1}{2} \sum_2^nk^2(k-1)=\frac{1}{2} \left(\sum_2^nk^3 -\sum_2^n k^2\right)$$

In genearl 
I can write (tell me if it's wrong or there is a more friendly way to write)
$$a_k=(-1)^{k+1}\sum_{\substack{I \subseteq \{1,\dotsb,n\} \\ |I|=k}}\prod_{i \in I}i $$
The questions are:


*

*Can I make this more explicit and simpler like the cases $a_1, a_2$?

*Can I write $a_k$ as sums of power of the first $n$ integers?

 A: First Consider the polynomial
$$P_n(x) = \prod_{k=1}^{n} (1+kx) := \sum_{j=0}^{n}a_j(n)x^j$$
Accounting the convention that $a_{j}(n) =0 $ if $j>n$ and notice this does not affect any of $P_n(x)'s$.
Then 

$$\boxed{ 1-P_n(-x)=1-(1-x)(1-2x)\dotsb (1-nx)} $$
  Therefore it suffices to find $a_j(n)$ $j= 0,1\cdots ,n$

On the other hand, we have 
$$ P_{n+1}(x) = \prod_{k=1}^{n+1} (1+kx) =  P_n(x)+(n+1)P_n(x)x := \sum_{j=0}^{n}a_j(n+1)x^j$$
But it easy to check that 
$$ P_n(x)+(n+1)P_n(x)x = 1+(n+1)a_n(n)x^{n+1}+ \sum_{j=1}^{n}[a_j(n)+(n+1)a_{j-1}(n)]x^j $$
By identification we get $a_{n+1}(n+1)=(n+1)a_{n}(n)$, $a_0(n)=1$ and 
$$ a_{j}(n+1)=a_j(n)+(n+1)a_{j-1}(n)~~\text{for}~~j= 1\cdots ,n$$

It is easy to see that $a_0(n)=1$, $a_1(n)=\binom{n+1}{2} = \frac{n(n+1)}{2}$ and $a_{n}(n) =n!$

Also, $\text{for}~~j= 1,\cdots ,n$
$$ a_{j}(n+1)-a_j(n)= (n+1)a_{j-1}(n)~$$
which leads to 
$$a_{j}(n+1)-a_j(0) = \sum_{k=1}^{n+1}a_{j}(k)-a_j(k-1) =\sum_{k=1}^{n+1}k a_{j-1}(k)   $$
Whence, $\text{for}~~j= 1,\cdots ,n$
$$a_{j}(n+1) = \sum_{k=1}^{n+1}k a_{j-1}(k)$$
