# 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.
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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 $P_n(k)=s(n+1,k).$

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This is a great answer and provides me with some context that I can work forward from. The reason I haven't accepted it is that it doesn't explicitly answer the bulleted points in my question. (e.g. given that $P_n(k)$ is a polynomial in $k$ what are the polynomials coefficients?). – Spencer Feb 9 '14 at 6:30


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