Help with this determinant I need help to calculate this determinant:
$$\begin{vmatrix}
1^k & 2^k & 3^k & \cdots & n^k \\
2^k & 3^k & 4^k & \cdots & (n+1)^k\\
3^k & 4^k & 5^k & \cdots & (n+2)^k\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
n^k & (n+1)^k & (n+2)^k & \cdots & (2n-1)^k
\end{vmatrix}$$
Where $2\leq n$ and $0\leq k \leq n-2.$
I did the $2\times 2$, $3\times 3$ and $4\times 4$ cases but I couldn't find a pattern to follow.
I did those cases by making zeros in the first column (except the $1^k$ in the first row) and then using the Laplace expansion.
The $2\times 2$ case equals: $3^k -4^k$, the  $3\times 3$ case equals: $15^k -16^k -20^k -27^k +24^k +24^k$

Another proof. 
This involves some tricks on manipulating polynomials. 
$\mathit{Proof}.\blacktriangleleft$
Consider
$$
f(z) = 
\begin{vmatrix}
(z+1)^k & (z+2)^k & (z+3)^k & \cdots & (z+n)^k\\
2^k & 3^k & 4^k & \cdots &  (n+1)^k\\
3^k & 4^k & 5^k & \cdots & (n+2)^k \\
\vdots & \vdots & \vdots & \ddots & \vdots\\
n^k & (n+1)^k & (n+2)^k & \cdots & (2n-1)^k
\end{vmatrix}.
$$
Then the computational definition determinant, i.e.
$$
\det(\boldsymbol A) = \sum_{\sigma \in \mathfrak S_n} \mathrm{sgn}(\sigma) \prod_1^n a_{\sigma(j), j}
$$ 
implies that $f(z)$ is a polynomial of degree $k$. The goal is to compute $f(0)$. Note that $f(1) = f(2) = \cdots = f(n-1) = 0$, since the $1^{\mathrm {st}}$ row coincides with the $ (j+1 )^{\mathrm {st}}$ row, and a determinant with duplicated rows is $0$. Therefore $f(z)$ is a real  polynomial with $n-1$ zeros while its degree is $k \leqslant n-2$. Thus the polynomial is just the zero polynomial, and the original determinant is $f(0) = 0$. $\blacktriangleright$
