A special Vandermonde system with integer coefficients Consider the Vandermonde system
  $$\left(\begin{matrix}  1&1&1&\dots&\dots&1\\
  1&2&3&4&\dots&n\\
  1&2^2&3^2&4^2&\dots&n^{2}\\
    1&2^3&3^3&4^3&\dots&n^{3}\\
    \vdots&&&&&\vdots\\
      1&2^{n-1}&3^{n-1}&4^{n-1}&\dots&n^{n-1}\\
  \end{matrix}
  \right)~
  \left(\begin{matrix} x_{1,n}\\ x_{2,n}\\\dots\\\vdots\\ \dots \\ x_{n,n}
  \end{matrix}\right)
  =\left(\begin{matrix} 1\\-1\\1\\-1\\ \vdots\\ (-1)^{n-1}
  \end{matrix}
  \right)
  $$
 The  solution for $n=6$ is  $(21, -70, 105, -84, 35, -6)$.
    Is there a formula for the general solution involving 2 parameters $j$ and $n $ and, may be, binomial coefficients?
 A: With help from OEIS A127717 we conjecture that
for $0\le p\lt n$
$$\sum_{k=1}^n k^p (-1)^{k+1}
\sum_{q=k}^n {q-1\choose k-1} q = (-1)^p.$$
This is
$$\sum_{q=1}^n q \sum_{k=1}^q {q-1\choose k-1}
(-1)^{k+1} k^p
= \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k}
(-1)^{k} (k+1)^p
\\ = \sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k}
(-1)^{k} p! [z^p] \exp((k+1)z)
\\ = p! [z^p] \exp(z)
\sum_{q=1}^n q \sum_{k=0}^{q-1} {q-1\choose k}
(-1)^{k}  \exp(kz)
\\ = p! [z^p] \exp(z)
\sum_{q=1}^n q (1-\exp(z))^{q-1}.$$
Now with $(1-\exp(z))^{q-1} = (-1)^{q-1} z^{q-1} + \cdots$ and $n\gt p$
we may certainly extend $q$ to infinity without contributing to $[z^p]$
and we find
$$p! [z^p] \exp(z)
\sum_{q\ge 1} q (1-\exp(z))^{q-1}
\\ = p! [z^p] \exp(z) \frac{1}{(1-(1-\exp(z)))^2}
\\ = p! [z^p] \exp(z) \exp(-2z) = p! [z^p] \exp(-z) = (-1)^p.$$
This is the claim. We see that the desired coefficient is given by
$$\bbox[5px,border:2px solid #00A000]{
x_{k,n} = (-1)^{k+1}
\sum_{q=k}^n {q-1\choose k-1} q.}$$ 
Remark. A careful examination of the OEIS entry shows that we
can simplify $x_{k,n}.$ We get for the sum
$$\sum_{q=k}^n {q-1\choose k-1} q
= k \sum_{q=k}^n {q\choose k}
= k \sum_{q=0}^{n-k} {q+k\choose k}
\\ = k \sum_{q=0}^{n-k} [z^k] (1+z)^{q+k}
= k [z^k] (1+z)^k \sum_{q=0}^{n-k} (1+z)^q
\\ = k [z^k] (1+z)^k \frac{(1+z)^{n-k+1}-1}{1+z-1}
\\ = k [z^{k+1}] (1+z)^k ((1+z)^{n-k+1}-1)
\\ = k [z^{k+1}] (1+z)^k (1+z)^{n-k+1}
= k [z^{k+1}] (1+z)^{n+1}.$$
We thus have
$$\bbox[5px,border:2px solid #00A000]{
x_{k,n} = (-1)^{k+1}
k {n+1\choose k+1}.}$$
