Vanishing of terms in power series expansion I am looking to find the power series expansion around $0$ of the rational function defined by $$f(z)=\prod_{l=0}^n (1-lz)^{(-1)^l {n\choose l}}.$$
By considering small $n$, I am led believe that the coefficients $a_r$ of the power series expansion satisfy $a_0=1$, $a_r=0$ for $0<r<n$, and $a_n=1+(-1)^{n-1}(n-1)! z^n$. I would like to see a quick proof of this statement (using any methods you like, the shorter the better), having tried e.g. residue theorem and induction in vain. 
Presumably this comes down to a certain identity between binomial coefficients?
 A: Partial Answer
I've got it down to a complicated binomial sum. 

Let $m_i=(-1)^i\binom{n}i$ be the exponent of the $i^\text{th}$ factor, for ease of notation. The following property of $m_i$ is almost certainly useful:
$$
\sum_{i=0}^ni^rm_i = 0\qquad\text{for }0<r<n\tag{1}
$$
Onto the problem at hand. Using the Leibniz rule for the $r^\text{th}$ derivative of a product of $n$ functions, 
$$
f^{(r)}(0) = \sum_{k_1+\dots+k_n=r}\binom{r}{k_1,k_2,\dots,k_n}\prod_{l=1}^n\frac{d^{k_i}}{dz^{k_i}}\bigg|_{z=0}(1-i z)^{m_i}
$$
Repeated differentiation introduces a falling factorial. Using Knuth's notation $x^{\underline n}=x(x-1)\cdots(x-n+1)$, we have
$$
\frac{d^{k_l}}{dz^{k_l}}\bigg|_{z=0}(1-i z)^{m_i}=(-i)^{k_l}(m_i)^{\underline{k_l}},
$$
so
$$
\boxed{f^{(r)}(0) = (-1)^r\!\!\!\!\!\sum_{k_1+\dots+k_n=r}\binom{r}{k_1,k_2,\dots,k_n}1^{k_1}(m_1)^{\underline{k_1}}\,2^{k_2}(m_2)^{\underline{k_2}}\cdots n^{k_n}(m_n)^{\underline{k_n}}}\tag{2}
$$
This looks suspiciously like the multinomial expansion of $(m_1+2m_2+\dots+nm_n)^r$. The problem is that some of the exponents are replaced by falling factorials. 

I've tried to expand this out for small values of $r$ (ignoring the $(-1)^r$ out front) and look for a pattern. When $r=1$, the sum is just $\sum_i im_i$, which by (1) is zero when $n>1$. When $r=2$, you get
$$
\sum_i i^2m_i(m_i-1) + \sum_{i\neq j}im_i jm_j=\left(\sum_i im_i\right)^2-\sum_i i^2m_i
$$
Applying $(1)$ twice, with $r=1,2$, the above is zero for $n>2$. Things get a little messier when $r=3$:
$$
\sum_{i}i^3 m_i(m_i-1)(m_i-2) + 3\sum_{i\neq j}i^2m_i(m_i-1)jm_j + \sum_{i\neq j\neq k\neq i}im_ijm_jkm_k
$$
We can collect all of the cubic terms nicely into $\left(\sum_i im_i\right)^3$, and the linear terms are $2\sum_{i} i^3m_i$. Both of these are $0$ by $(1)$ when $n>3$. The quadratic terms are
$$
-3\sum_i i^3m_i^2-3\sum_{i\neq j}i^2m_ijm_j=-3\sum_{i,j} i^2m_ijm_j=-3\left(\sum_i i^2m_i\right)\left(\sum_j jm_j\right)
$$
which is again $0$ for $n>3$. 

To summarize these results, let $s_r = \sum_i i^rm_i$. We have shown
$$
\begin{align}
-f^{(1)}(0) &= s_1\\
f^{(2)}(0) &= s_1^2 - s_2\\
-f^{(3)}(0) &= s_1^3 - 3s_1s_2 + 2s_3
\end{align}
$$
The signed Stirling numbers of the first kind are appearing; these resemble the expansion $x^{\underline n}=\sum_k s(n,k)x^k$ of the falling factorial into powers of $k$. No doubt a little more elbow grease will expose a general pattern which can be proved...
Addendum: Ok, I can hazard a guess. Notice the sum of the coefficients in each case is $r!$, and the indices of each summand are a partition of $r$, such that the coefficient is the number of permtuations with that cycle structure. This leads to the following guess:
For any permutation $\pi\in S_r$, let $\text{cyc}(\pi)$ the set of the lengths of its cycle lengths. 
$$
\boxed{f^{(r)}(0)\stackrel{?}{=} (-1)^r\sum_{\pi\in S_r}\prod_{c\in \text{cyc}(\pi)}(-1)^{c+1}s_{c}}
$$
If the above were true, your guesses about $a_n$ for $1\le r\le n$ would easily follow.
A: Work with the logarithm, then exponentiate.
$$\log(f(z))= \sum_{m=0}^{n} (-1)^m\binom{n}{m}\log(1-m\,z)=\sum_{m=0}^n(-1)^{m+1}\binom{n}{m} \sum_{k=1}^{\infty}\frac{(m\,z)^k}{k}.$$
Exchange summations in the last expression; first one has upper finite summation limit, so its O.K. 
$$\log(f(z))=  -\sum_{k=1}^{\infty}\frac{z^k}{k}\sum_{m=0}^n(-1)^{m}\binom{n}{m} m^k.$$
It is well known (differentiate the binomial theorem) that the inside sum is 0 for $k<n$ and has the value $(-1)^n\,n!$ for $k=n.$ Thus
$\log(f(z))= -(-1)^n\,z^n(n-1)!+O(z^{n+1}).$  Exponentiation gives the result of the OP's that
$$f(z) = 1  +z^n(n-1)!(-1)^{n-1}+O(z^{n+1}), $$
