Coefficients of $(x-1)(x-2)\cdots(x-k)$ I'm interested in the coefficients of $x$ in the expansion of,
$$ (x-1)(x-2)\cdots(x-k) = x^k + P_1(k) x^{k-1} + P_2(k)x^{k-2} + \cdots + P_k(k),$$
Where $k$ is an integer. In particular I am interested in thinking of these coefficients as polynomials in $k$. 
Its not to hard to show that,
$$ P_1(k) = -\sum_{i=1}^k i =-k(k-1)/2 $$
 $$ P_2(k) = \sum_{i=2}^k i \sum_{j=1}^{i-1} j = k^4/8 + k^3/12-k^2/8-k/12$$
And I am pretty sure that $$ P_n(k) = (-1)^k\sum_{i_1=n}^k i_1 \sum_{i_2=1}^{i_1-1}i_2 \sum_{i_3=1}^{i_2-1}i_3\cdots i_{n-1}\sum_{i_n=1}^{i_{n-1}-1}i_n$$
I haven't gotten around to proving it but it works for $P_1$, $P_2$ and $P_3$ which gives me some confidence in the formula. For the purpose of this question assume that the formula works in general. 
The last polynomial is a bit awkward because $P_k(k) = (-1)^kk!$ meaning that the coefficients are heavily dependent upon $k$ and somewhat ill defined. However I am primarily interested in $P_n$ when $n<k$.
My questions are the following,


*

*Is there a simple formula for the coefficients of $P_n(k)$.

*Is there a tight upper bound $M_k \geq P_n(x)$ for $x=1,2,\ldots,k$ which holds for all $n$. 

*I would also be interested in an upper bound on the coefficients if their explicit form is unavailable.

 A: It is the Stirling  numbers of the first kind. 

By definition they are the coefficients in the expansion
$(x)_n = \sum_{k=0}^n s(n,k) x^k,$
where $(x)_n$ is the falling factorial
$(x)_n = x(x-1)(x-2)\cdots(x-n+1).$

So  $P_n(k)=s(n+1,k).$
A: $\newcommand{\+}{^{\dagger}}%
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$$
\sum_{k = 0}^{n}a_{k}x^{k} = \sum_{k = 0}^{n}\left[k!\,a_{k}\right]\,{x^{k} \over k!} 
$$
Then,
$$
a_{k}
=
\left.{1 \over k!}\totald[k]{\pars{\sum_{k = 0}^{n}a_{k}x^{k}}}{x}
\right\vert_{x\ =\ 0}
$$
In your case, the 'simple formula' you are looking for is given by:
$$
P_{i}
=
\left.
{1 \over \pars{k - i}!}\totald[k - i]{\bracks{\pars{x - 1}\ldots\pars{x - k}}}{x}
\right\vert_{x\ =\ 0}
$$
